6. THERMOCHEMISTRY. Solutions to Exercises

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1 6. THERMOCHEMISTRY Solutions to Eercises Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multiple-step problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 6.1 Substitute into the formula E k = 1/ mv using SI units: E k = 1/ kg ( m/s) = = J J 1 cal J = = cal 6. Heat is evolved; therefore, the reaction is eothermic. The value of q is kj. 6.3 The thermochemical equation is N H 4 (l) + N O 4 (l) 3N (g) + 4H O(g); H = kj H equals kj for 1 mol N H 4 and for mol N H a. N H 4 (l) + 1/N O 4 (l) 3/N (g) + H O(g); H = kj b. 4H O(g) + 3N (g) N H 4 (l) + N O 4 (l); H = 1049 kj

2 10 CHAPTER The reaction is N H 4 (l) + N O 4 (l) 3N (g) + 4H O(g); H = kj 1 mol N 10.0 g N H 4 H g 1 mol N O 4 mol NH kj 4 1 mol NH 4 = = -164 kj 6.6 Substitute into the equation q = s m t to obtain the heat transferred. The temperature change is t = t f - t i = C C = 80.0 C Therefore, q = s m t = 0.449J/(g C) 5.00 g 80.0 C = = J 6.7 The total mass of the solution is obtained by adding the volumes together and by using the density of water (1.000 g/ml). This gives = 75 ml, or 75 g. The heat absorbed by the solution is q = s m t = J/(g C) 75 g (31.8 C C) = J The heat released by the reaction, q rn, is equal to the negative of this value, or J. To obtain the enthalpy change for the reaction, you need to calculate the moles of HCl that reacted. This is Mol HCl = 1.0 mol/l L = mol The enthalpy change for the reaction can now be calculated. H = J mol = J/mol = -54 kj/mol Epressing this result as a thermochemical equation, HCl(aq) + NaOH(aq) NaCl(aq) + H O(l); H = -54 kj

3 THERMOCHEMISTRY Use Hess's law to find H for 4Al(s) + 3MnO (s) Al O 3 + 3Mn(s) from the following data for reactions 1 and : Al(s) + 3/O (g) Al O 3 (s); H = kj (reaction 1) Mn(s) + O (g) MnO (s); H = -50 kj (reaction ) If you take reaction 1 and multiply it by, you obtain 4Al(s) + 3O (g) Al O 3 (s); H = (-1676 kj) = -335 kj Since the desired reaction has three MnO on the left side, reverse reaction and multiply it by three. The result is 3MnO (s) 3Mn(s) + 3O (g) H = -3 (-50 kj) = 1560 kj If you add the two reactions and corresponding enthalpy changes, you obtain the enthalpy change of the desired reaction 4Al(s) + 3O (g) Al O 3 (s) H = -335 kj 3MnO (s) 3Mn(s) + 3O (g) H = 1560 kj 4Al(s) +3MnO (s) Al O 3 (s) + 3Mn(s) H = -179 kj 6.9 The vaporization process, with the H f values given below the substances, is H O(l) H O(g) (kj) The calculation is H o vap o o = Σ n H (products) - Σ m H (reactants) f f = H o [H f O(g)] - o H f [H O(l)] = (-41.8 kj) - (-85.8 kj) = 44.0 kj

4 1 CHAPTER The reaction, with the H f values given below the substances, is 3NO (g) + H O(l) HNO 3 (aq) + NO(g) (kj) The calculation is o H rn o o = Σ n H (products) - Σ m H (reactants) f f o o o o = [ H (HNO3) + H (NO)] - [3 H (NO) + H (HO)] f f = [(-07.4) + (90.9)] kj - [3(33.10) + (-85.8)] kj = = kj f f 6.11 The net chemical reaction, with the H f values given below the substances, is NH + 4 (aq) + OH - (aq) NH 3 (g) +H O(l) (-13.5) (-30.0) (-45.90) (-85.8) (kj) The calculation is o H rn o o o = [ H (NH3) + H (HO)] - [ H (NH 4+ o ) + H (OH - )] f f f f = [(-45.90) + (-85.8)] - [(-13.5) + (-30.0)] = = 61.6 kj Answers to Concept Checks 6.1 The photovoltaic cells collect the sun s energy, converting it to electric energy. This electric energy is stored in the battery as chemical energy, which is later changed back to electric energy that runs a motor. As the motor rotates, it changes the electric energy to kinetic energy (energy of motion) of the motor, then of water, which in turn is changed to potential energy ( energy of position) of water as the water moves upward in the gravitational field of earth. 6. a. This reaction is the one shown in the problem, and it has a positive H, so the reaction is endothermic. b. This reaction is simply twice reaction a, so it is also endothermic. c. This reaction is the reverse of reaction a, so it is eothermic. (continued)

5 THERMOCHEMISTRY 13 d. This reaction is simply twice that of reaction c, so it is more eothermic than c. Thus, d is the most eothermic reaction. 6.3 You can think of the sublimation of ice as taking place in two stages. First, the solid melts to liquid, then the liquid vaporizes. The first process has an enthalpy, H fus. The second process has an enthalpy, H vap. Therefore, the total enthalpy, which is the enthalpy of sublimation, is the sum of these two enthalpies: H sub = H fus + H vap. Answers to Review Questions 6.1 Energy is the potential or capacity to move matter. Kinetic energy is the energy associated with an object by virtue of its motion. Potential energy is the energy an object has by virtue of its position in a field of force. Internal energy is the sum of the kinetic and potential energies of the particles making up a substance. 6. In terms of SI base units, a joule is kg m /s. This is equivalent to the units for the kinetic energy (½mv ) of an object of mass m (in kg) moving with speed v (in m/s). 6.3 Originally, a calorie was defined as the amount of energy required to raise the temperature of one gram of water by one degree Celsius. At present, the calorie is defined as J. 6.4 At either of the two highest points above the earth in a pendulum's cycle, the energy of the pendulum is all potential energy and is equal to the product mgh (m = mass of pendulum, g = constant acceleration of gravity, and h = height of pendulum). As the pendulum moves downward, its potential energy decreases from mgh to near zero, depending on how close it comes to the earth's surface. During the downward motion, its potential energy is converted to kinetic energy. When it reaches the lowest point (middle) of its cycle, the pendulum has its maimum kinetic energy and minimum potential energy. As it rises above the lowest point, its kinetic energy begins to be converted to potential energy. When it reaches the other high point in its cycle, the energy of the pendulum is again all potential energy. By the law of conservation of energy, this energy cannot be lost, only converted to other forms. At rest, the energy of the pendulum has been transferred to the surroundings in the form of heat. 6.5 As the heat flows into the gas, the gas molecules gain energy and move at a faster average speed. The internal energy of the gas increases.

6 14 CHAPTER An eothermic reaction is a chemical reaction or a physical change in which heat is evolved (q is negative). For eample, burning one mol of methane, CH 4 (g), yields carbon dioide, water, and kj of heat. An endothermic reaction is a chemical reaction or physical change in which heat is absorbed (q is positive). For eample, the reaction of one mol of barium hydroide with ammonium nitrate absorbs kj of heat in order to form ammonia, water, and barium nitrate. 6.7 Changes in internal energy depend only on the initial and final states of the system, which are determined by variables such as temperature and pressure. Such changes do not depend on any previous history of the system. 6.8 The enthalpy change equals the heat of reaction at constant pressure. 6.9 At constant pressure, the enthalpy change is positive (the enthalpy increases) for an endothermic reaction It is important to give the states when writing an equation for H because H depends on the states of all reactants and products. If any state changes, H changes When the equation for the reaction is doubled, the enthalpy is also doubled. When the equation is reversed, the sign of H is also reversed. 6.1 First, convert the 10.0 g of water to moles of water, using its molar mass (18.0 g/mol). Net, using the equation, multiply the moles of water by the appropriate mole ratio (1 mol CH 4 / mol H O). Finally, multiply the moles of CH 4 by the heat of the reaction ( kj/mol CH 4 ) The heat capacity (C) of a substance is the quantity of heat needed to raise the temperature of the sample of substance one degree Celsius (or one kelvin). The specific heat of a substance is the quantity of heat required to raise the temperature of one gram of a substance by one degree Celsius (or one kelvin) at constant pressure A simple calorimeter consists of an insulated container (for eample, a pair of styrene coffee cups as in Figure 6.1) with a thermometer. The heat of the reaction is obtained by conducting the reaction in the calorimeter. The temperature of the miture is measured before and after the reaction. The heat capacity of the calorimeter and its contents also must be measured Hess s law states that, for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps. In other words, no matter how you go from reactants to products, the enthalpy change for the overall chemical change is the same. This is because enthalpy is a state function No, you can still obtain the enthalpy for the desired reaction. You will need to come up with a system of reactions that can be combined to give the desired reaction and know the enthalpy changes for each of the steps. Then, using Hess s law, you can obtain the enthalpy change for the reaction under study.

7 THERMOCHEMISTRY The thermodynamic standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: one atm pressure and the specified temperature (usually 5 C) The reference form of an element is the most stable form (physical state and allotrope) of the element under standard thermodynamic conditions. The standard enthalpy of formation of an element in its reference form is zero The standard enthalpy of formation of a substance, H f, is the enthalpy change for the formation of one mole of the substance in its standard state from its elements in their reference form and in their standard states. 6.0 The equation for the formation of H S(g) is H (g) + 1/8S 8 (rhombic) H S(g) 6.1 The reaction of C(g) + 4H(g) CH 4 (g) is not an appropriate equation for calculating the H f of methane because the most stable form of each element is not used. Both H (g) and C(graphite) should be used instead of H(g) and C(g), respectively. 6. A fuel is any substance that is burned or similarly reacted to provide heat and other forms of energy. The fossil fuels are petroleum (oil), gas, and coal. They were formed millions of years ago when aquatic plants and animals were buried and compressed by layers of sediment at the bottoms of swamps and seas. Over time this organic matter was converted by bacterial decay and pressure to fossil fuels. 6.3 One of the ways of converting coal to methane involves the water-gas reaction. C(s) + H O(g) CO(g) + H (g) In this reaction, steam is passed over hot coal. This miture is then reacted over a catalyst to give methane. CO(g) + 3H (g) CH 4 (g) + H O(g) 6.4 Some possible rocket fuel/oidizer combinations are H /O and hydrazine/dinitrogen tetroide. The chemical equations for their reactions are H (g) + 1/O (g) H O(g); H = -4 kj N H 4 (l) + N O 4 (l) 3N (g) + 4H O(g); H = kj

8 16 CHAPTER 6 Answers to Conceptual Problems 6.5 Kinetic energy is proportional to mass and to speed squared. Compare the kinetic energy of the smaller car with that of the larger car (twice the mass) assuming both are traveling at the same speed. The larger car would have twice the kinetic energy of the smaller car. Or, we could say that the smaller car has only half the kinetic energy of the larger car. Now suppose the speed of the smaller is increased by a factor of two (so it is now moving at twice its original speed). Its kinetic energy is increased by a factor of four. Therefore, the smaller car now has one-half times four times, or twice, the kinetic energy of the larger car. The smaller car has the greater kinetic energy. 6.6 The equation says that one mol of butane reacts with (13/) mol of oygen to yield four mol carbon dioide and five mol water. The reaction yields a certain amount of heat, which you can symbolize as q. So a yields heat q. On the other hand, b is only one mol oygen, not (13/) mol. So, b yields heat equal to (/13) q. This result might be easier to see by first looking at c. Note that the one mole of carbon dioide stated in c is only one-fourth that given in the equation. This means that c yields one-fourth q (just the inverse of the coefficient of the equation). Similarly, d yields one-fifth q. Therefore, b yields the least heat. 6.7 a. After the water is placed into the freezer, it will lose heat to the freezer, so q sys is negative. b. The water will have turned to ice. c. The initial enthalpy (of the water) is higher than the final enthalpy (of the ice). d. After several hours, the temperature of the water will be -0 C. 6.8 a. Heat will flow from the iron block to the aluminum block. Fe Al b. The aluminum block will be absorbing heat, so q sys will be positive. c. The temperatures of the aluminum block and the iron block will both be the same. d. The heat gained by the aluminum will equal the heat lost by the iron. Each heat term will be equal to mass sp ht T. This gives (0.0 g)(0.901 J/g C)(T f C) = (0.0 g)(0.449 J/g C)(50.0 C - T f ) (continued)

9 THERMOCHEMISTRY 17 You can cancel the mass and the units from both sides to give a simplified epression (0.901)(T f C) = (0.449)(50.0 C - T f ) This can be simplified to give T f = 6.995, or T f = = 46.7 C. 6.9 You can imagine this process taking place in two steps: first, the preparation of water vapor from the elements, and second, the change of the vapor to liquid. Here are the equations: H (g) + ½O (g) H O(g); H f H O(g) H O(l); - H vap The last equation is the reverse of the vaporization of water, so the enthalpy of the step is the negative of the enthalpy of vaporization. The enthalpy change for the preparation of one mole of liquid water, H, is the sum of the enthalpy changes for these two steps: H = H f + (- H vap ) = H f - H vap 6.30 The epression for the heat is q = s m t. For the same amount of heat and mass, the product s t must be constant. The metal with the smaller specific heat will have the larger t. Since the specific heat for aluminum (0.901 J/g C) is larger than for iron (0.449 J/g C), the block of iron will have the larger t and be warmer a. The heat lost by the metal is equal to the heat gained by the water. Since q = s m t, the heat gained by the water is directly proportional to t. Since t is larger for metal A, it lost more heat. Now, each metal has the same mass and t, so the specific heat is directly proportional to q. Since q is larger for A, the specific heat is larger for A. b. The metal with the higher specific heat will have absorbed more heat to reach the starting temperature of 95 o C; therefore, it will release more heat to the water causing the water to reach a higher temperature. The beaker with metal A will rise to the higher temperature. 6.3 Silver metal reacts with a halogen to produce the corresponding silver halide. For eample, silver reacts with fluorine to produce silver fluoride. Each reaction corresponds to the reaction forming the silver halide, so you look in the table of enthalpies of formation of compounds (Table 6.). The most eothermic reaction would be the one with the most negative enthalpy of formation. That would be the reaction for the formation of silver fluoride.

10 18 CHAPTER Let us write H rn for the enthalpy change when one mole of P 4 S 3 burns in O to give P 4 O 10 and SO. In principle, you could calculate H rn from enthalpies of formation for the reactants and products of this reaction. You would require the values for P 4 S 3, O (which equals zero), P 4 O 10, and SO. Enthalpies of formation for the products, P 4 O 10 and SO, are given in Table 6. in the tet This means that, if you have measured H rn, you can use the enthalpies of formation of P 4 O 10 and SO to calculate the enthalpy of formation for P 4 S 3. What you have done is this: you have used enthalpies of combustion to calculate enthalpies of formation. This is the idea most often used to obtain enthalpies of formation a. Since H is positive, the reaction is endothermic, and the solution will be colder. b. While the salt dissolves, heat will flow into the beaker to raise the temperature of the water back to the initial temperature. c. After the water returns to room temperature, q for the system will be zero. Solutions to Practice Problems Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multiple-step problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off The heat released, in kcal, is kj 1000 J 1 kj 1 cal J 1 kcal 1000 cal = = kcal 6.36 The heat released, in kcal, is kj 1000 J 1 kj 1 cal J 1 kcal 1000 cal = = kcal

11 THERMOCHEMISTRY The kinetic energy, in J, is E k = 1/ lb kg 57 mi 1609 m 1 lb 1 h 1 mi 1 h 3600 s The kinetic energy, in calories, is = = J E k = J 1 cal J = = cal 6.38 The kinetic energy, in J, is E k = 1/ 35 gr kg 50 ft 1 yd 1 gr 1 s 3 ft m 1 yd The kinetic energy, in cal, is = 449 = J E k = 449 J 1 cal J = 1073 = cal 6.39 To insert the mass of one molecule of ClO in the formula, multiply the molar mass by the reciprocal of Avogadro's number. The kinetic energy in J is g 1 mol 1 kg 306 m E k = 1/ 1 mol molec g 1 s = = J/molec To insert the mass of one molecule of N O in the formula, multiply the molar mass by the reciprocal of Avogadro's number. The kinetic energy in J is 44.0 g 1 mol 1 kg 379 m E k = 1/ 1 mol molec g 1 s = = J/molec.

12 0 CHAPTER Endothermic reactions absorb heat, so the sign of q will be positive because energy must be gained by the system from the surroundings. The flask will feel cold to the touch. 6.4 Eothermic reactions evolve heat, so the sign of q will be negative because energy is lost from the system to the surroundings. The flask will feel hot to the touch The gain of 66. kj of heat per two mol NO means the reaction is endothermic. Because energy is gained by the system from the surroundings, q is positive and is +66. kj for two mol of NO reacting The release of 939 kj of heat per two mol HCN means the reaction is eothermic. Because energy is lost from the system to the surroundings, q is negative, and because the reaction involves two mol HCN, q for the reaction is -939 kj The reaction of Fe(s) with HCl must yield H and FeCl 3. To balance the hydrogen, HCl must be written first as a reactant. However, to balance the chlorine, 3 HCl must be written finally as a reactant, and FeCl 3 must be written as a product: Fe(s) + 6HCl(aq) FeCl 3 (aq) + 3H (g) To write a thermochemical equation, the sign of H must be negative because heat is evolved: Fe(s) + 6HCl(aq) FeCl 3 (aq) + 3H (g); H = kj = kj 6.46 The decomposition of two mol KClO 3 to KCl and O must yield two mol KCl to balance the potassium and chlorine. To balance the oygen, three mol O must be written as a product: KClO 3 (s) KCl(s) + 3O (g) To write a thermochemical equation, the sign of H must be negative because heat is evolved: KClO 3 (s) KCl(s) + 3O (g); H = kj

13 THERMOCHEMISTRY The first equation is P 4 (s) + 5O (g) P 4 O 10 (s); H = kj The second equation is P 4 O 10 (s) P 4 (s) + 5O (g); H =? The second equation has been obtained by reversing the first equation. Therefore, to obtain H for the second equation, H for the first equation must be reversed in sign: -(-3010) = kj The first equation is CS (l) + 3O (g) CO (g) + SO (g); H = kj The second equation is 1/CS (l) + 3/O (g) 1/CO (g) + SO (g); H =? The second equation has been obtained by dividing each coefficient in the first equation by two. Therefore, to obtain H for the second equation, H for the first equation must be divided by two: (-1077) = = -539 kj The first equation is 1/4P 4 O 10 (s) + 3/H O(l) H 3 PO 4 (aq); H = -96. kj The second equation is P 4 O 10 (s) + 6H O(l) 4H 3 PO 4 (aq); H =? The second equation has been obtained from the first by multiplying each coefficient by four. Therefore, to obtain the H for the second equation, the H for the first equation must be multiplied by four: = kj 6.50 The first equation is 4NH 3 (g) + 5O (g) 4NO(g) + 6H O(g); H = -906 kj The second equation is NO(g) + 3/H O(g) NH 3 (g) + 5/4O (g); H =? The second equation has been obtained by reversing the first equation and dividing each coefficient by four. Therefore, to obtain H for the second equation, the H for the first equation must be reversed in sign and divided by four: = 6.50 = 7 kj.

14 CHAPTER Because nitric oide is written as NO in the equation, the molar mass of NO equals g per mol NO. From the equation, two mol NO evolve 114 kj heat. Divide the 114 kj by the g/mol NO to obtain the amount of heat evolved per gram of NO: -114 kj mol NO 1 mol NO g NO = = kj g NO 6.5 Because hydrogen is written as H in the equation, the molar mass of H equals.016 g per mol H. From the equation, two mol H evolve 484 kj heat. Divide this by the.016 g/mol H to obtain the amount of heat evolved per gram of hydrogen: -484 kj mol H 1 mol H.016 g H = = -10 kj g H 6.53 The molar mass of ammonia is g/mol. From the equation, four moles of NH 3 evolve 167 kj of heat. Divide 35.8 g NH 3 by its molar mass and the four moles of NH 3 in the equation to obtain the amount of heat evolved: 35.8 g NH 3 1 mol NH g NH -167 kj mol NH = = kj 6.54 The molar mass of H S is g/mol. From the equation, two moles of H S evolve 1036 kj of heat. This gives 1 mol H 6.7 g H S S g HS kj mol HS = = -406 kj 6.55 The molar mass of C 3 H 8 is g/mol. From the equation, one mol C 3 H 8 evolves 043 kj heat. This gives -369 kj 1 mol C3H8-043 kj g C3H 8 1 mol C3H8 = = 7.96 g C 3 H The molar mass of C H 5 OH is g/mol. From the equation, one mol C H 5 OH evolves 135 kj heat. This gives -93 kj 1 mol CH5OH -135 kj g CH5OH 1 mol C H OH = 10.9 = 10.9 g C H 5 OH 5

15 THERMOCHEMISTRY Multiply the 180 g (0.180 kg) of water by the specific heat of 4.18 J/(g C) and by t to obtain heat in joules: 180 g (96 C - 19 C) 4.18 J 1 g C = = J 6.58 Multiply the g (1.63 kg) iron by the specific heat of J/(g C) and by t to obtain heat in joules: g (178 C - 1 C) J 1 g C = = J 6.59 Use the J/g (.6 kj/g) heat of vaporization to calculate the heat of condensation. Then, use it to calculate t, the temperature change. Heat of condensation = J 1 g 168 g = J Temperature change = t = J g 1 g C J = = 5.81 C 6.60 Use the 334 J/g (0.334 kj/g) heat of fusion to calculate the heat of melting. Then, use the heat of melting to calculate t, the temperature change. Heat of melting = 38.0 g 334 J 1 g = 169 J Heat melting + heat warming = heat lost by 10 g H O Let T = the final temperature, and substitute into the above equation: 169 J J 38.0 g (T - 0 C) 1 g C = 4.18 J 1 g C 10 g (1.0 C - T) Solve the above equation for T. T = J J ( ) J/ C = = 5.54 C

16 4 CHAPTER The enthalpy change for the reaction is equal in magnitude and opposite in sign to the heat-energy change occurring from the cooling of the solution and calorimeter. q calorimeter = (1071 J/ C)(1.56 C C) = J Thus, 15.3 g NaNO 3 is equivalent to J heat energy. The amount of heat absorbed by mol NaNO 3 is calculated from J (opposite sign): mol NaNO g NaNO3 1 mol NaNO J 15.3 g NaNO = J 3 Thus, the enthalpy change, H, for the reaction is J, or 0.5 kj, per mol NaNO The enthalpy change for the reaction is equal in magnitude and opposite in sign to the heat energy produced from the warming of the solution and the calorimeter. q calorimeter = (158 J/ C)(38.7 C C) = J Thus, 3.6 g CaCl is equivalent to J heat energy. The amount of heat released by mol CaCl is calculated from J (opposite sign): g CaCl mol CaCl 1 mol CaCl J 3.6 g CaCl = J Thus, the enthalpy change, H, for the reaction is J, or kj, per mol CaCl The energy change for the reaction is equal in magnitude and opposite in sign to the heat energy produced from the warming of the solution and the calorimeter. q calorimeter = (9.63 kj/ C)( C) = kj Thus,.84 g C H 5 OH is equivalent to kj heat energy. The amount of heat released by mol C H 5 OH is calculated from kj (opposite sign): 1.00 mol C H 5 OH g CH5OH 1 mol CH5OH kj.84 g CH5OH = kj Thus, the enthalpy change, H, for the reaction is kj/mol ethanol.

17 THERMOCHEMISTRY The energy change for the reaction is equal in magnitude and opposite in sign to the heat energy produced from the warming of the solution and the calorimeter. q calorimeter = (1.05 kj/ C)(37.18 C C) = kj Thus, 3.51 g C 6 H 6 is equivalent to kj heat energy. The amount of heat released by mol C 6 H 6 is calculated from kj (opposite sign): 1.00 mol C 6 H g C6H 6 1 mol C6H kj 3.51 g C6H 6 = Thus, the enthalpy change, H, for the reaction is kj/mol benzene Using the equations in the data, reverse the direction of the first reaction, and reverse the sign of its H. Then multiply the second equation by two, multiply its H by two, and add. Setup: N (g) + H O(l) N H 4 (l) + O (g); H = (-6. kj) (-1) H (g) + O (g) H O(l); H = (-85.8 kj) () N (g) + H (g) N H 4 (l); H = 50.6 kj 6.66 Using the equations in the data, reverse the direction of the first reaction, and reverse the sign of its H. Then divide the second equation by two, divide its H by two, and add. Setup: H O(l) + 1/O (g) H O (l); H = (-98.0 kj) (-1) H (g) + 1/O (g) H O(l); H = ( kj) () H (g) + O (g) H O (l); H = kj 6.67 Using the equations in the data, multiply the second equation by two, and reverse its direction; do the same to its H. Then multiply the first equation by two and its H by two. Finally, multiply the third equation by three and its H by three. Then add. Setup: 4NH 3 (g) N (g) + 6H (g); H = (-91.8 kj) (-) N (g) + O (g) 4NO(g); H = (180.6 kj) () 6H (g) + 3O (g) 6H O(g); H = ( kj) (3) 4NH 3 (g) + 5O (g) 4NO(g) + 6H O(g); H = kj

18 6 CHAPTER Using the equations in the data, reverse the direction of the second equation and reverse the sign of its H. Then, reverse the direction of the first equation and multiply by 1/; do the same to its H. Finally, multiply the third equation and its H by 1/. Add all three equations. Setup: CH 4 (g) C(graph) + H (g); H = (-74.9 kj) (-1) NH 3 (g) 1/N (g) + 3/H (g); H = (-91.8 kj) (-1/) 1/H (g) + C(graph) + 1/N (g) HCN(g); H = (70.3 kj) (1/) CH 4 (g) + NH 3 (g) HCN(g) + 3H (g); H = = 56.0 kj 6.69 After reversing the first equation in the data, add all the equations. Setup: C H 4 (g) + 3O (g) CO (g) + H O(l); H = (-1411 kj) CO (g) + 3H O(l) C H 6 (g) + 7/O (g); H = (-1560) (-1) H (g) + 1/O (g) H O(l); H = (-86 kj) C H 4 (g) + H (g) C H 6 (g); H = -137 kj 6.70 After reversing the first equation, double the second and third equations, and then add. Setup: CO (g) + H O(l) CH 3 COOH(l) + O (g); H = (-874 kj) (-1) C(graph) + O (g) CO (g); H = (-394 kj) () H (g) + O (g) H O(l); H = (-86 kj) () C(graph) + H (g) + O (g) CH 3 COOH(l); H = -486 kj 6.71 Write the H values (Appendi C) underneath each compound in the balanced equation: C H 5 OH(l) C H 5 OH(g) (kj) H vap = [ H f (C H 5 OH)] - [ H f (C H 5 OH)] = [-35.4] - [-77.7] = +4.3 kj 6.7 Write the H values (Table 6.) underneath each compound in the balanced equation: CCl 4 (l) CCl 4 (g) (kj) H vap = [ H f (CCl 4 )] - [ H f (CCl 4 )] = [-95.98] - [-135.4] kj = kj

19 THERMOCHEMISTRY Write the H values (Table 6.) underneath each compound in the balanced equation: H S(g) + 3O (g) H O(l) + SO (g) (-0.50) 3(0) (-85.8) (-96.8) (kj) H = Σ n H (products) - Σ m H (reactants) = [(-85.8) + (-96.8)] - [(-0.50) + 3(0)] kj = = -114 kj 6.74 Write the H values (Table 6.) underneath each compound in the balanced equation: CS (l) + 3O (g) CO (g) + SO (g) (0) (-96.8) (kj) H = Σ n H (products) - Σ m H (reactants) = [(-393.5) + (-96.8)] - [(87.90) + 3(0)] = kj 6.75 Write the H values (Appendi C) underneath each compound in the balanced equation: Fe O 3 (s) + 3CO(g) Fe(s) + 3CO (g) (-110.5) (0) 3(-393.5) (kj) H = Σ n H (products) - Σ m H (reactants) = [(0) + 3(-393.5)] - [(-85.5) + 3(-110.5)] = -3.5 kj 6.76 Write the H values (Appendi C) underneath each compound in the balanced equation: PbS(s) + 3O (g) SO (g) + PbO(s) (-98.3) 3(0) (-96.8) (-19.4) (kj) H = Σ n H (products) - Σ m H (reactants) = [(-96.8) + (-19.4)] - [(-98.3) + 3(0)] = = kj

20 8 CHAPTER Write the H values (Table 6.) underneath each compound in the balanced equation: HCl(g) H + (aq) + Cl - (aq) (kj) H = Σ n H (products) - Σm H (reactant) = [(0) + (-167.)] - [-9.31] = = kj 6.78 Write the H values (Table 6.) underneath each compound in the balanced equation: CaCO 3 (s) + CO (g) + H O(l) Ca + (aq) + HCO 3 - (aq) (-69.0) (kj) H = Σ n H (products) - Σ m H (reactants) = [(-54.8) + (-69.0)] - [(-106.9) + (-393.5) + (-85.8)] = kj 6.79 Calculate the molar heat of formation from the equation with the H f values below each substance; then convert to the heat for 10.0 g of MgCO 3 using the molar mass of MgCO 3 (s) MgO(s) + CO (g) kj H = Σ n H (products) - Σ m H (reactants) H = kj + ( kj) - ( kj) = kj Heat = 10.0 g 1 mol 84.3 g kj mol = = 13.9 kj 6.80 Calculate the molar heat of formation from the equation with the H f values below each substance; then convert to the heat for 10.0 g of BaCO 3 using the molar mass of BaCO 3 (s) BaO(s) + CO (g) kj H = Σ n H (products) - Σ m H (reactants) H = kj + ( kj) - ( kj) = 74.7 kj Heat = 10.0 g 1 mol g 74.7 kj mol = 13.9 = 13.9 kj

21 THERMOCHEMISTRY 9 Solutions to General Problems 6.81 The SI units of force must be kg m/s (= newton, N) to be consistent with the joule, the SI unit of energy: kg m s m = kg m s = joule, J 6.8 Solving E p = mgh, you obtain g = E p /mh. Thus, the SI unit of g is (kg m /s )/(kg m) = m/s Using Table 1.5 and J/cal, convert the 686 Btu/lb to J/g: 686 Btu 1 lb 5 cal 1 Btu J 1 cal 1 lb kg 1 kg 3 10 g = = J/g 6.84 Using Table 1.5 and J/cal, convert the Btu/lb to J/g: Btu 1 lb 5 cal 1 Btu J 1 cal 1 lb kg 1 kg 3 10 g = = J/g 6.85 Substitute into the equation E p = mgh, and convert to SI units. E p = 1.00 lb m s 167 ft kg 1 lb m 3 ft = kg s m = 6 J At the bottom, all the potential energy is converted to kinetic energy, so E k = 5.8 kg m /s. Because E k = 1/mv, solve for v, the speed (velocity): Speed = E k 1/ m = kg m / s 1/ 1.00 lb kg/lb = = 31.6 m/s

22 30 CHAPTER Substitute into the equation E k = 1/mv, and convert to SI units. E k = 1/ 354 lb kg 1 lb km 10 m s 1 km = kg m /s = J 6.87 The equation is CaCO 3 (s) CaO(s) + CO (g); H = kj Use the molar mass of g/mol to convert the heat per mol to heat per 7.3 g. 1.3 g CaCO 3 1 mol CaCO g CaCO kj 1 mol CaCO 3 = = 37.9 kj 6.88 The equation is CaO(s) + H O(l) Ca(OH) (s); H = -65. kj Use the molar mass in the conversion. 4.5 g CaO 1 mol CaO g CaO -65. kj 1 mol CaO = = -8.5 kj The heat released is 8.5 kj The equation is HCHO (l) + O (g) CO (g) + H O(l) Use the molar mass of g/mol to convert kj/5.48 g to H per mol of acid kj 5.48 g HCHO g HCHO 1 mol HCHO = = -55 kj/mol

23 THERMOCHEMISTRY The equation is HC H 3 O (l) + O (g) CO (g) + H O(l) Use the molar mass of g/mol to convert -5.0 kj/3.58 g to H per mol of acid kj 3.58 g HC H O g HCH3O 1 mol HCHO 3 3 = -87. = -87 kj/mol 6.91 The heat gained by the water at the lower temperature equals the heat lost by the water at the higher temperature. Each heat term is mass sp ht T. This gives (54.9 g)(4.184 J/g C)(T f C) = (1.0 g)(4.184 J/g C)(5.7 C - T f ) The specific heat and the units can be canceled from both sides to give (54.9)(T f C) = (1.0)(5.7 C - T f ) After rearranging, you get 75.9 T f = This gives T f = = 37.4 C. 6.9 The heat gained by the water at the lower temperature equals the heat lost by the water at the higher temperature. Each heat term is mass sp ht T. This gives (45.4 g)(4.184 J/g C)(T f C) = (0.5 g)(4.184 J/g C)(66. C - T f ) The specific heat and the units can be canceled from both sides to give (45.4)(T f C) = (0.5)(5.7 C - T f ) After rearranging, you get 65.9 T f = This gives T f = = 45. C Divide the 35 J heat by the mass of lead and the t to obtain the specific heat. Specific heat = 35 J 11.6 g (35.5 C C) = = 0.18 J g C 6.94 Divide the 47.0 J heat by the mass of copper and the t to obtain the specific heat. Specific heat = 47.0 J 35.4 g (3.45 C) = = J g C

24 3 CHAPTER The energy used to heat the Zn comes from cooling the water. Calculate q for water: q wat = specific heat mass t q wat = 4.18 J 50.0 g (96.68 C C) = J g C The sign of q for the Zn is the reverse of the sign of q for water because the Zn is absorbing heat: q met = -(q wat ) = -( ) = J Specific heat = J 5.3 g (96.68 C C) = = J g C 6.96 The energy given up by the metal is used to heat the water. Calculate q for water: q wat = specific heat mass t q wat = 4.18 J 6.7 g (30.00 C C) = J g C The sign of q for the metal is the reverse of the sign of q for water because the metal is giving up heat: q met = -(q wat ) = J Specific heat = J 19.6 g (30.00 C C) = = J g C 6.97 First, multiply the molarities by the volumes, in liters, to get the moles of NaOH and the moles of HCl, and thus determine the limiting reactant. Mol NaOH = M V = M L = mol Mol HCl = M V = M L = mol Therefore, NaOH is the limiting reactant. The total volume of the system is 14.1 ml ml = 46.4 ml. Since the density of water is 1.00 g/ml, the total mass of the system is 46.4 g. Net, set the heat released by the reaction (mol enthalpy of reaction) equal to the heat absorbed by the water (m sp ht. T). This gives ( mol)( J/mol) = (46.4 g)(4.184 J/g C)(T f C) After dividing, this gives T f C = C, or T f = 5.63 = 5.6 C

25 THERMOCHEMISTRY First, multiply the molarities by the volumes, in liters, to get the moles of KOH and the moles of HBr and thus determine the limiting reactant. Mol KOH = M V = 1.05 M L = mol Mol HBr = M V = 1.07 M L = mol Therefore, HBr is the limiting reactant. The total volume of the system is 9.1 ml ml = 50.0 ml. Since the density of water is 1.00 g/ml, the total mass of the system is 50.0 g. Net, set the heat released by the reaction (mol enthalpy of reaction) equal to the heat absorbed by the water (m sp ht. T). This gives ( mol)( J/mol) = (50.0 g)(4.184 J/g C)(T f C) After dividing, this gives T f C = C, or T f = 7.76 = 7.8 C 6.99 Use t and the heat capacity of 547 J/ C to calculate q: q = C t = (547 J/ C)( ) C = J (6.378 kj) Energy is released in the solution process in raising the temperature, so H is negative: H = kj 6.48 g LiOH 3.95 g LiOH 1 mol LiOH = = -3.6 kj/mol Use t and the heat capacity of 68 J/ C to calculate q: q = C t = (68 J/ C)( ) C = J ( kj) Energy is absorbed in the solution process in lowering the temperature, so the sign of H must be reversed, making the heat positive: H = kj 1.45 g KNO g KNO3 1 mol KNO 3 3 = = 34.9 kj/mol Use t and the heat capacity of kj/ C to calculate q: q = C t = (13.43 kj/ C)( ) C = kj As in the previous two problems, the sign of H must be reversed, making the heat negative: H = kj g HC H O g HCH3O 1 mol HCHO 3 3 = = kj/mol

26 34 CHAPTER Use t and the heat capacity of 15.8 kj/ C to calculate q: q = C t = (15.8 kj/ C)( ) C = 8.53 kj As in the previous three problems, the sign of H must be reversed, making the heat negative: H = kj g sugar g sugar 1 mol sugar = = kj/mol Using the equations in the data, reverse the direction of the first reaction, and reverse the sign of its H. Then add the second and third equations and their H's. H O(g) + SO (g) H S(g) + 3/O (g); H = (-518 kj) (-1) H (g) + 1/O (g) H O(g); H = (-4 kj) 1/8S 8 (rh.) + O (g) SO (g); H = (-97 kj) H (g) + 1/8S 8 (rh.) H S(g); H = -1 kj Using the equations in the data, reverse the direction of the reaction involving oidation of the glycol, and reverse the sign of its H. Then add half of the reaction involving oidation of C H 4 O, and half of its H. CO (g) + 3H O(l) HOC H 4 OH(l) + 5/O (g); H = ( kj) (-1) C H 4 O(g) + 5/O (g) CO (g) + H O(l); H = (-61. kj) (1/) C H 4 O(g) + H O(l) HOC H 4 OH(l); H = kj Write the H values (Table 6.) underneath each compound in the balanced equation. CH 4 (g) + H O(g) CO(g) + 3H (g) (0) (kj) H = [ (0)] - [(-74.87) + (-41.8)] = = 06. kj

27 THERMOCHEMISTRY Write the H values (Table 6.) underneath each compound in the balanced equation. CH 4 (g) + O (g) CO(g) + 4H (g) (-74.87) 0 (-110.5) 4(0) (kj) H = [(-110.5) + 4(0)] - [(-74.87) + 0] = = kj Write the H values (Table 6.) underneath each compound in the balanced equation. The H f of -635 kj/mol CaO is given in the problem. CaCO 3 (s) CaO(s) + CO (g) (kj) H = [(-635.1) + (-393.5)] - [-106.9] = = 178 kj Write the H values (Table 6.) underneath each compound in the balanced equation. NaHCO 3 (s) Na CO 3 (s) + H O(g) + CO (g) (-950.8) (kj) H = [( ) + (-41.8) + (-393.5)] - [(-950.8)] = kj Calculate the molar heat of reaction from the equation with the H f values below each substance; then convert to the heat for the reaction at 5 C. H (g) + O (g) H O(l) 0.0 kj 0.0 kj kj H = (-85.8 kj) = kj The moles of oygen in.000 L with a density of 1.11 g/l is: 1.11 g O.000 L O 1 L O 1 mol O 3.0 g O = mol O The heat of reaction from mol O is: mol O kj mol O = = kj

28 36 CHAPTER Calculate the molar heat of reaction from the equation with the H f values below each substance; then convert to the heat for the reaction at 5 C. Cl (g) + Na(s) NaCl(s) 0.0 kj 0.0 kj kj H = ( kj) = -8. kj The moles of chlorine in L Cl with a density of.46 g/l is:.46 g Cl 1 mol Cl L Cl = mol Cl L Cl 70.9 g Cl The heat of reaction from mol Cl is: -8. kj mol Cl = = -114 kj mol Cl Let H f = the unknown standard enthalpy of formation for sucrose. Then write this symbol and the other H f values underneath each compound in the balanced equation, and solve for H f using the H of kj for the reaction. C 1 H O 11 (s) + 1O (g) 1CO (g) + 11H O(l) H f 1(0) 1(-393.5) 11(-85.8) (kj) H = = [1(-393.5) + 11(-85.8)] - [ H f + 1(0)] H f = -5 kj/mol sucrose 6.11 Let H f = the unknown standard enthalpy of formation for acetone. Then write this symbol and the other H f values underneath each compound in the balanced equation, and solve for H f using the H of kj for the reaction. CH 3 COCH 3 (l) + 4O (g) 3CO (g) + 3H O(l) H f 4(0) 3(-393.5) 3(-85.8) (kj) H = = [3(-393.5) + 3(-85.8)] - [ H f + 4(0)] H f = = -47 kj/mol acetone

29 THERMOCHEMISTRY First, calculate the heat evolved from the molar amounts represented by the balanced equation. From Appendi C, obtain the individual heats of formation, and write those below the reactants and products in the balanced equation. Then, multiply the molar heats of formation by the number of moles in the balanced equation, and write those products below the molar heats of formation. Al(s) + 3NH 4 NO 3 (s) 3N (g) + 6H O(g) + Al O 3 (s) 0.0 kj kj 0.0 kj kj (kj/mol) 0.0 kj kj 0.0 kj kj (kj/eqn.) The total heat of reaction of two moles of Al and three moles of NH 4 NO 3 is H = ( kj) = kj Now, 45 kj represents the following fraction of the total heat of reaction: 45 kj 09.7 kj = Thus, 45 kj requires the fraction of the moles of each reactant: , or , mol of Al and , or , mol of NH 4 NO 3. The mass of each reactant and the mass of the miture are as follows: mol Al 6.98 g/mol Al = g of Al mol NH 4 NO g/mol NH 4 NO 3 = g NH 4 NO g Al g NH 4 NO 3 = = 35.5 g of the miture First, calculate the heat evolved from the molar amounts represented by the balanced equation. From Appendi C, obtain the individual heats of formation, and write those below the reactants and products in the balanced equation. Then, multiply the molar heats of formation by the number of moles in the balanced equation, and write those products below the molar heats of formation. Al(s) + Fe O 3 (s) Fe(l) + Al O 3 (s) 0.0 kj kj kj kj (kj/mol) 0.0 kj kj kj kj (kj/eqn.) The total heat of reaction of two moles of Al and one mole of Fe O 3 is H = (-85.5 kj) = kj (continued)

30 38 CHAPTER 6 Now, 348 kj represents the following fraction of the total heat of reaction: 348 kj kj = Thus, 348 kj requires the fraction of the moles of each reactant: 0.416, or 0.843, mol of Al and , or 0.416, mol of Fe O 3. The mass of each reactant and the mass of the miture are as follows: mol Al 6.98 g/mol Al =.750 g Al mol Fe O g/mol Fe O 3 = g Fe O g Al g Fe O 3 = = 90.1 g of the miture Solutions to Cumulative-Skills Problems The heat lost, q, by the water(s) with a temperature higher than the final temperature must be equal to the heat gained, q, by the water(s) with a temperature lower than the final temperature. q(lost by water at higher temp) = q(gained by water at lower temp) Σ (s m t) = Σ (s m t) Divide both sides of the equation by the specific heat, s, to eliminate this term, and substitute the other values. Since the mass of the water at 50.0 C is greater than the sum of the masses of the other two waters, assume the final temperature will be greater than 37 C and greater than 15 C. Use this to set up the three t epressions, and write one equation in one unknown, letting t equal the final temperature. Simplify by omitting the grams from 45.0 g, 5.0 g, and 15.0 g. Σ (m t) = Σ (m t) 45.0 (50.0 C - t) = 5.0 (t C) (t - 37 C) 50 C t = 5.0 t C t C t = C t = = 37.4 C (the final temperature)

31 THERMOCHEMISTRY The heat lost, q, by the substances(s) with a temperature higher than the final temperature must be equal to the heat gained, q, by the substances(s) with a temperature lower than the final temperature. q(lost) = q(gained) Σ (s m t) = Σ (s m t) Since the temperature of the iron, 95.0 C is far larger than the temperatures of the other two substances, assume the final temperature will be greater than 35.5 C and greater than 5 C. Then, write the heat term for the iron on the left and the sum of the heat terms for water and ethanol on the right. Since the masses, m, are equal, divide both sides by m to eliminate this term, and solve one equation in one unknown, letting t equal the final temperature. (To simplify the setup, only the temperature unit is retained.) Σ (s t) = Σ (s t) (95.0 C - t) =.43 (t C) (t C) C t =.43 t C t C t = C t = = 33.1 C (the final temperature) The final temperature is actually less than that of the ethanol, but rewriting the equation to account for this still gives the same final temperature First, calculate the mole fraction of each gas in the product, assuming 100 g product: Mol CO = 33 g CO 1 mol CO/8.01 g CO = mol CO Mol CO = 67 g CO 1 mol CO /44.01 g CO = 1.5 mol CO Mol frac. CO = mol CO ( ) mol = mol CO Mol frac. CO = ( ) mol = Now, calculate the starting moles of C, which equal the total moles of CO and CO : Starting mol C(s) = 1.00 g C 1 mol C/1.01 g C = mol C = mol CO + CO (continued)

32 40 CHAPTER 6 Use the mol fractions to convert mol CO + CO to mol CO and mol CO : Mol CO = Mol CO = mol CO 1 mol total mol CO 1 mol total mol total = mol CO mol total = mol CO Now, use enthalpies of formation (Table 6.) to calculate the heat of combustion for both: C(s) + 1/O (g) CO(g) ecess (mol) (kj/mol) (kj/ mol) C(s) + O (g) CO (g) ecess (mol) (kj/mol) (kj/ mol) Total H = (-18.47) = -.48 = - kj Heat released = kj The balanced equations are CH 4 (g) + O (g) CO (g) + H O(l) C H 6 (g) + 3.5O (g) CO (g) + 3H O(l) First, calculate the moles of CH 4 and C H 6 (let 80.0% = g, and 0.0% = 0.00 g). 1 mol CH Mol CH 4 = g CH g CH = mol CH 4 4 Mol C H 6 = 0.00 g C H 6 1 mol CH g C H = mol C H 6 6 (continued)

33 THERMOCHEMISTRY 41 Now, use the enthalpies of formation (Table 6.) to calculate the total heat: CH 4 (g) + O (g) CO (g) + H O(l) ecess ( ) (mol) (kj/mol) (kj/eqn.) C H 6 (g) + 3.5O CO (g) + 3H O(l) ecess ( ) 3( ) (mol) (kj/mol) (kj/eqn.) For the combustion of CH 4 : H = [(-8.51) + (-19.6)] - [(-3.734) + 0] = kj For the combustion of C H 6 : H = [(-5.34) + (-5.703)] - [(-0.563) + 0] = kj For the combustion of the total (1.00 g) of CH 4 and C H 6 : H = ( ) = = kj The heat evolved = 54.8 kj The H for the reactions to produce CO and CO in each case is equal to the H f of the respective gases. Thus H to produce CO is kj/mol, and H to produce CO is kj/mol. Since one mol of graphite is needed to produce one mol of either CO or CO, one equation in one unknown can be written for the heat produced using m for the moles of CO and (.00 - m) for moles of CO kj = m ( kj) + (.00 - m) ( kj) = m m mol CO = m = 1.081; mass CO = g/mol = 30.6 = 30.3 g mol CO = (.00 - m) = 0.919; mass CO = mol 44.0 g/mol = 40.4 = 40. g

34 4 CHAPTER The H values for each reaction are as follows: CH 4 (g) + O (g) CO (g) + H O(l) H = kj + (-85.8 kj) - ( kj) - 0 = kj C H 4 (g) + 3O (g) CO (g) + H O(l) H = ( kj) + (-85.8 kj) - (5.47 kj) - 0 = kj Since 10.0 g of CH 4 and C H 4 produce 50 kj of heat, one equation in one unknown can be written for the heat produced using m for the mass of CH 4 and ( m) for the mass of C H 4. Use m /16.0 for the moles of CH 4 and ( m) /8.0 for the moles of C H kj = (m /16.0) ( kj) + [( m) /8.0] ( kj) -50 = m + ( ) m mass CH 4 = m = = 3.06 = 3.1 g mass percentage CH 4 = (3.06 g 10.0 g) 100% = 3.06 = 3.1% 6.11 The equation is 4NH 3 (g) + 5O (g) 4NO(g) + 6H O(g); H = -906 kj First, determine the limiting reactant by calculating the moles of NH 3 and of O ; then, assuming one of the reactants is totally consumed, calculate the moles of the other reactant needed for the reaction g NH 3 1 mol NH g NH = mol NH mol O 0.0 g O 3.00 g O = mol O 1 mol O mol O 3.00 g O = mol NH 3 needed Because NH 3 is present in ecess of what is needed, O must be the limiting reactant. Now calculate the heat released on the basis of the complete reaction of mol O : (continued)

35 THERMOCHEMISTRY kj H = mol O 5 mol O = = kj The heat released by the complete reaction of the 0.0 g (0.650 mol) of O (g) is 113 kj. 6.1 The equation is CS (g) + 3Cl (g) S Cl (g) + CCl 4 (g); H = -3 kj First, determine the limiting reactant by calculating the moles of CS and of Cl ; then, assuming one of the reactants is totally consumed, calculate the moles of the other reactant needed for the reaction. 1 mol CS 10.0 g CS g CS 1 mol Cl 10.0 g Cl g Cl = mol CS = mol Cl 1 mol CS mol Cl 3 mol Cl = mol CS Because CS is present in ecess of what is needed, Cl must be the limiting reactant. Now, calculate the heat released on the basis of the complete reaction of mol Cl : H = -30 kj 3 mol Cl mol Cl = = kj The heat released by the complete reaction of 10.0 g of Cl (g) is 10.8 kj The equation is N (g) + 3H (g) NH 3 (g); H = kj a. To find the heat evolved from the production of 1.00 L of NH 3, convert the 1.00 L to mol NH 3 using the density and molar mass (17.03 g/mol). Then convert the moles to heat ( H) using H : (continued)