Directions: Answer the following 1. When writing a complete ionic equation, a. what types of substances should be shown as dissociated/ionized? soluble ionic compounds, acids, bases b. What types of substances should not be shown as dissociated/ionized? most molecular, insoluble ionic 2. How does electrical potential energy change between particles of like charge as the distance between them increases? As the distance between particles increases, the repulsive force (potential energy) decreases. How does electrical potential energy change between particles of opposing charge as the distance between them increases? As the distance between particles increases, the attractive force (potential energy) decreases. 3. Differentiate between an open, closed and isolated thermodynamic system. Isolated system: no exchange of heat, work, or matter with the surroundings; Closed system: exchange of heat and work, but not matter with the surroundings; Open system: exchange of heat, work and matter with the surroundings. Directions: Fill in the table for each of the variables. definition meaning if positive meaning if negative ΔE Change in internal energy (E f -E i ) The system is gaining energy from the environment from work or heat. The system is releasing energy to the environment from work or heat. w Work Work is being done on the system, so internal energy increases. Work is being done by the system, so internal energy decreases. q Heat Heat is being gained by the system, so internal energy increases. Heat is being lost by the system, so internal energy decreases. ΔH Change in enthalpy (H f -H i ) Enthalpy increasing products are higher in enthalpy than reactants. Enthalpy decreasing products are lower in enthalpy than reactants ΔT Change in temperature (T f -T i ) Temperature increasing. Temperature decreasing ΔV Change in volume (V f -V i ) Volume increasing. Volume decreasing 4. Describe what is happening in system where q = -51 kj and w = +25 kj. Heat is being released by the system (q is negative) and work is being done on the system (w is positive) 5. Describe what is happening in system where q = +321 kj and w = +132 kj. Heat is being added by the system (q is positive) and work is being done on the system (w is positive) 6. Define a state function. A property that depends only on the condition or "state" of the system, and not on the path used to obtain the current conditions. Energy, enthalpy, temperature, volume, pressure, and temperature are examples of state functions; heat and work are examples of non-state functions. 7. What is pressure-volume (PV) work? Pressure-volume work (or PV work) occurs when the volume V of a system changes. PV work is often measured in units of litre-atmospheres where 1L atm = 101.325J. 8. In the formula, w = -PΔV, why is PΔV a negative value? because this formula describes work at a constant pressure, if ΔV is negative (volume is shrinking) work is being done on the system (squeezing the gas) and w must be positive the negative sign makes the w value positive; if the gas is expanding it is doing work on the environment (volume is increasing) and ΔV would be positive the negative sign would make the w value negative 9. In what types of reactions would PV work require special attention? Reactions in which gas is produced or consumed, thereby changing the pressure of the system
10. What are the conditions of the enthalpy statement ΔH o = -117 kj? 1 atm and 25 o C 11. Why is the enthalpy of formation of an element in its standard state equal to zero? The standard enthalpy of formation of a substance is defined as the enthalpy change when 1 mol of the substance is formed from the most stable form of its elements in their standard states at a pressure of 1 atm and a temperature of 25 C. All elements in their standard states (for example, oxygen in the form of O 2,C in the form of graphite, H 2, I2, F2) have a standard enthalpy of formation of zero, because there is no change involved when they are formed from themselves. 12. Describe an endothermic reaction in terms of the reactants and products. The energy necessary to break the reactants is greater than the energy released by the products in their formation. 13. Describe an exothermic reaction in terms of the reactants and products. The energy released in the formation of the products is greater than the energy necessary to break the reactants. 14. Why is the sign of ΔH for a reaction reversed when the reaction direction is reversed? The amount of energy involved in the formation of a substance is equal but opposite in sign to the energy required to decompose the susbstance. 15. How are heat capacity and specific heat related? Different? Heat capacity and specific heat are both measures of how much energy is required to raise the temperature of a substance by 1 o C (or 1K). Specific heat applies to 1 gram of the substance as opposed to the the entire sample of the substance. 16. Why does ΔH = q for most reactions carried out in a calorimeter? ΔH = q for the conditions that a chemist usually works with: a constant pressure environment, and the system does no other work and has no other work done on it than the work of expansion. 17. How are coffee-cup calorimetry and bomb calorimetry alike? Different? Both types of calorimetry are used to determine heat flow resulting from a change to a system. They capitalize on the known heat capacity of a substance that absorbs or releases heat in the case of coffee cup calorimetry, that substance is usually water; in a bomb calorimeter, the substance is the bomb and surrounding water. Bomb calorimetry is constant volume meaning no PV work can be done since the volume of the bomb cannot change. Coffee cup calorimetry is constant pressure meaning the pressure is held constant (this is usually used when reactions are carried out in solution). 18. Why does the formula for calculating heat flow in a bomb calorimeter not include mass? In bomb calorimetry the heat capacity of the bomb is determined experimentally in order to calibrate it. The mass of the bomb is constant and incorporated into the heat capacity C cal. 19. In the bomb calorimetry formula, q = -C cal ΔT, why is C cal ΔT a negative value? The q value represents the heat of the reaction. When the temperature of the bomb increases, the ΔT is positive, and the q value must be negative as the bomb has heated due to energy being released by the reaction. The negative sign accomplishes the q rxn = -q sur that is used in coffee cup calorimetry. 20. Describe Hess Law. The total enthalpy of a reaction is independent of the reaction pathway. This means that if a reaction is carried out in a series of steps, the enthalpy change (DH) for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps. 21. Define enthalpy of formation. change in enthalpy involved in forming 1 mole of a substance from its constituent elements 22. The symbol ΔH o f means standard enthalpy of formation (heat of formation at standard conditions, 1atm & 25 o C) 23. Why does standard enthalpy of formation equal zero for pure elements in their most stable form? It requires no energy to form an element in its most stable elemental form. This is sort of its natural resting position as an element.
Directions: Calculate each of the following in the space provided. 24. If 357.0 atm L of work are done on a 64.00 ml volume of gas at a constant pressure of 2.20 atm, what is the new volume of gas? w = -P V w = 357. atm L P = 2.20 atm V = w/-p = 357 atm L / -2.20 atm = -162 L (-162000 ml) V f = V i + V = 64.00mL + (-162000 ml) = Oops. You get a negative volume of gas. This one won t work the way it is written. How about this instead: If 0.0357 atm L of work are done on a 64.00 ml volume of gas at a constant pressure of 2.20 atm, what is the new volume of gas? w = -P V w = 0.0357 atm L P = 2.20 atm V = w/-p = 0.0357 atm L / -2.20 atm = -0.0162 L (-16.2 ml) V f = V i + V = 64.00mL + (-16.2 ml) = 47.8 ml The volume decreases because work in being done on the volume of gas. 25. Compute the enthalpy change for the production of 17.1 g of Fe 2 O 3 using the following unbalanced equation. FeO(s) + O 2 (g) Fe 2 O 3 (s) ; H o = -540. kj 4FeO(s) + O 2 (g) 2Fe 2 O 3 (s) ; H o = -540. kj 17.1g? kj 17.1 g Fe 2 O 3 1 mol Fe 2 O 3-540. kj 159.687 g Fe 2 O 3 2 mol Fe 2 O 3 = -28.9 kj 26. A 46.2-g sample of copper is heated to 95.4 C and then placed in a calorimeter containing 70.0 g water at 19.6 C. After the metal cools, the final temperature of metal and water is 21.8 C. Calculate the specific heat capacity of copper, assuming that all the heat lost by the copper is gained by the water. q = mc T First calculate the energy transferred to the water using the values for the water: q w =? m w = 70.0 g c w = 4.184 J/(g o C) T i(w) = 19.6 o C T f(w) = 21.8 o C T w = 21.8 o C 19.6 o C = 2.2 o C q w = 70.0 g 4.184 J/(g o C) 2.2 o C = 644.336 J q m = -q w (because the water gained the energy directly and only from the metal) Calculate the specific heat of the metal using values for the metal: q m = -644.336 J m m = 46.2 g c w =? T i(m) = 95.4 o C T f(m) = 21.8 o C T w = 21.8 o C 95.4 o C = -73.6 o C c = q m /(m m T m ) = -644.336 J / (46.2 g -73.6 o C) = 0.189 J/(g o C) 27. A 0.1964-g sample quinone (C 6 H 4 O 2 ) is burned in a bomb calorimeter that has a heat capacity of 1.56 kj/ C. The temperature of the calorimeter increases by 3.2 C. Calculate the energy of combustion of quinone in joules per gram and per mole. q rxn = -C cal ΔT
q rxn =? C cal = 1.56 kj/ C ΔT = 3.2 C q rxn = -(1.56 kj/ C 3.2 C) = -5.0 kj Convert kj to J: -5.0 kj = -5.0 x 10 3 J Finding q rxn per gram: -5.0 x 10 3 J / 0.1964 g = -25000 J/g Converting q rxn per gram to q rxn per mole: -25000 kj 108.096 g C 6 H 4 O 2 g 1 mol C 6 H 4 O 2 = -2.7 x 10 6 J / mol 28. Find the H for the reaction below, given the following reactions and subsequent H values: HCl(g) + NaNO 2 (s) HNO 2 (l) + NaCl(s) ½ reverse 2 NaCl(s) + H 2 O(l) 2 HCl(g) + Na 2 O(s) ΔH = 507 kj ½ reverse NO(g) + NO 2 (g) + Na 2 O(s) 2 NaNO 2 (s) ΔH = 427 kj ½ forward NO(g) + NO 2 (g) N 2 O(g) + O 2 (g) ΔH = 43 kj ½ reverse 2 HNO 2 (l) N 2 O(g) + O 2 (g) + H 2O(l) ΔH = 34 kj HCl(g) + ½ Na 2 O(s) NaCl(s) + ½ H 2 O(l) NaNO 2 (s) ½ NO(g) + ½ NO 2 (g) + ½ Na 2 O(s) ½ NO(g) + ½ NO 2 (g) ½ N 2 O(g) + ½ O 2 (g) ½ N 2 O(g) + ½ O 2 (g) + ½ H 2 O(l) HNO 2 (l) ΔH = - 253.5 kj ΔH = 213.5 kj ΔH = - 21.5 kj ΔH = - 17 kj ΔH rxn -78 kj Directions: Using a table of enthalpies of formation (like the one on page 184), calculate the following: 29. The reusable booster rockets of the space shuttle use a mixture of aluminum and ammonium perchlorate as fuel. A possible reaction is: Calculate ΔH for this reaction. 3Al(s) + 3NH 4 ClO 4 (s) Al 2 O 3 (s) + AlCl 3 (s) + 3NO(g) + 6H 2 O(g) H o = [sum of enthalpies of products] [sum of enthalpies of reactants] H o = [H f (Al 2 O 3 (s)) + H f (AlCl 3 (s)) + 3 H f (NO(g)) + 6 H f (H 2 O(g))] [3 H f (NH 4 ClO 4 (s))] Using values from the worksheet 05.7: H o = [-1675.7 kj + -704.2 kj + 3(90.4 kj) + 6(-241.8 kj)] [3(-295 kj)] = -2674 kj 30. Consider the reaction: 2ClF 3 (g) + 2NH 3 (g) N 2 (g) + 6HF(g) + Cl 2 (g); ΔH = 1196 kj Calculate ΔH o f for ClF 3 (g). H o = [sum of enthalpies of products] [sum of enthalpies of reactants] H o = [6 H f (HF(g))] [2 H f (ClF 3 (g)) + 2 H f (NH 3 (g))] ΔH = -1196 kj Using values from the worksheet 05.7:
Let x stand in for H f (ClF 3 (g)) (We re supposed to be using ΔH to find a missing enthalpy of formation.) -1196 kj = [6(-268.6 kj)] [2x + 2(-46.2 kj)] (Watch distributing the - here, especially on the 2(-46.2 kj) ) -1196 kj = -1611.6 kj 2x + 92.4 kj, solve for x x = -162 kj H o f(clf 3 ) = -162 kj