A calorimeter is to be calibrated: 72.55 g of water at 71.6 C added to a calorimeter containing 58.85 g of water at 22.4 C. After stirring and waiting for the system to equilibrate, the final temperature reached 47.3 C. Calculate the heat capacity of the calorimeter. (The specific heat capacity of water is 4.184 J g 1 C 1 ). 1) Energy lost by the hot water: q = m C p ΔT q = (72.55 g) (4.184 J/g 1 C 1 ) (24.3 C) q = 7376.24 J 2) Energy gained by the cold water: q = m C p ΔT q = (58.85 g) (4.184 J/g 1 C 1 ) (24.9 C) q = 5818.54 J 3) The calorimeter got the rest: 7376.24 5818.54 = 1557.7 J 4) Heat capacity of the calorimeter: 1557.7 J / 24.9 C = 62.558 J/ C (round off as you see fit) A student wishes to determine the heat capacity of a coffee cup calorimeter. After mixing 100.0 g of water at 58.5 C with 100.0 g of water, already in the calorimeter, at 22.8 C, the final temperature of the water is 39.7 C. Calculate the heat capacity of the calorimeter in J/ C. (Use 4.184 J/g C as the specific heat of water.) 1) Heat given up by warm water: q = (100.0 g) (18.8 C) (4.184 J/g C) = 7865.92 J 2) Heat absorbed by water in the calorimeter: q = (100.0 g) (16.9 C) (4.184 J/g C) = 7070.96 J 3) The difference was absorbed by the calorimeter: 7865.92 7070.96 = 794.96 J 4) Calorimeter constant: 794.96 J / 16.9 C = 47.0 J/ C
Germain Henri Hess, in 1840, discovered a very useful principle which is named for him: The enthalpy of a given chemical reaction is constant, regardless of the reaction happening in one step or many steps. Another way to state Hess' Law is: If a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation equals the sum of the enthalpy changes of the other chemical equations. Example #1: Calculate the enthalpy for this reaction: 2C(s) + H 2 (g) > C 2 H 2 (g) ΔH =??? kj Given the following thermochemical equations: C 2 H 2 (g) + (5/2)O 2 (g) > 2CO 2 (g) + H 2 O(l) C(s) + O 2 (g) > CO 2 (g) H 2 (g) + (1/2)O 2 (g) > H 2 O(l) ΔH = 1299.5 kj ΔH = 393.5 kj ΔH = 285.8 kj 1) Determine what we must do to the three given equations to get our target equation: a) first eq: flip it so as to put C 2 H 2 on the product side b) second eq: multiply it by two to get 2C c) third eq: do nothing. We need one H 2 on the reactant side and that's what we have. 2) Rewrite all three equations with changes applied: 2CO 2 (g) + H 2 O(l) > C 2 H 2 (g) + (5/2)O 2 (g) 2C(s) + 2O 2 (g) > 2CO 2 (g) H 2 (g) + (1/2)O 2 (g) > H 2 O(l) ΔH = +1299.5 kj ΔH = 787 kj ΔH = 285.8 kj Notice that the ΔH values changed as well. 3) Examine what cancels: 2CO 2 first & second equation H 2 O first & third equation (5/2)O 2 first & sum of second and third equation 4) Add up ΔH values for our answer: +1299.5 kj + ( 787 kj) + ( 285.8 kj) = +226.7 kj
Example #2: Calculate the enthalpy of the following chemical reaction: Given: CS 2 (l) + 3O 2 (g) > CO 2 (g) + 2SO 2 (g) C(s) + O 2 (g) > CO 2 (g) ΔH = 393.5 kj/mol S(s) + O 2 (g) > SO 2 (g) ΔH = 296.8 kj/mol C(s) + 2S(s) > CS 2 (l) ΔH = +87.9 kj/mol 1) What to do to the data equations: leave eq 1 untouched (want CO 2 as a product) multiply second eq by 2 (want to cancel 2S, also want 2SO 2 on product side) flip 3rd equation (want CS 2 as a reactant) 2) The result: C(s) + O 2 (g) > CO 2 (g) ΔH = 393.5 kj/mol 2S(s) + 2O 2 (g) > 2SO 2 (g) ΔH = 593.6 kj/mol < note multiply by 2 on the ΔH CS 2 (l) > C(s) + 2S(s) 3) Add the three revised equations. C and 2S will cancel. 4) Add the three enthalpies for the final answer. ΔH = 87.9 kj/mol < note sign change on the ΔH Example #3: Given the following data: SrO(s) + CO 2 (g) > SrCO 3 (s) 2SrO(s) > 2Sr(s) + O 2 (g) 2SrCO 3 (s) > 2Sr(s) + 2C(s, gr) + 3O 2 (g) ΔH = 234 kj ΔH = +1184 kj ΔH = +2440 kj Find the ΔH of the following reaction: C(s, gr) + O 2 (g) > CO 2 (g) 1) Analyze what must happen to each equation: a) first eq flip it (this put the CO 2 on the right hand side, where we want it) b) second eq do not flip it, divide through by two (no flip because we need to cancel the SrO, divide by two because we only need to cancel one SrO) c) third equation flip it (to put the SrCO 3 on the other side so we can cancel it), divide
by two (since we need to cancel only one SrCO 3 ) Notice that what we did to the third equation also sets up the Sr to be cancelled. Why not also multiply first equation by two (to get 2SrO for canceling)? Because we only want one CO 2 in the final answer, not two. Notice also that I ignored the oxygen. If everything is right, the oxygen will take care of itself. 2) Apply all the above changes (notice what happens to the ΔH values): SrCO 3 (s) > SrO(s) + CO 2 (g) SrO(s) > Sr(s) + (1/2)O 2 (g) Sr(s) + C(s, gr) + (3/2)O 2 (g) > SrCO 3 (s) ΔH = +234 kj ΔH = +592 kj ΔH = 1220 kj 3) Here is a list of what gets eliminated when everything is added: SrCO 3, SrO, Sr, (1/2)O 2 The last one comes from (3/2)O 2 on the left in the third equation and (1/2)O 2 on the right in the second equation. 4) Add the equations and the ΔH values: +234 + (+592) + ( 1220) = 394 C(s, gr) + O 2 (g) > CO 2 (g) ΔH f = 394 kj Notice the subscripted f. This is the formation reaction for CO 2. Example #4: Given the following information: 2NO(g) + O 2 (g) > 2NO 2 (g) 2N 2 (g) + 5O 2 (g) + 2H 2 O(l) > 4HNO 3 (aq) N 2 (g) + O 2 (g) > 2NO(g) ΔH = 116 kj ΔH = 256 kj ΔH = +183 kj Calculate the enthalpy change for the reaction below: 3NO 2 (g) + H 2 O(l) > 2HNO 3 (aq) + NO(g) ΔH =??? 1) Analyze what must happen to each equation: a) first eq flip; multiply by 3/2 (this gives me 3NO 2 as well as the 3NO which will be necessary to get one NO in the final answer)
b) second eq divide by 2 (give me my two nitric acid in the final answer) c) third eq flip (cancels 2NO as well as nitrogen) 2) Comment on the oxygens: a) step 1a above puts (3/2)O 2 on the right b) step 1b puts (5/2)O 2 on the left c) step 1c puts (2/2)O 2 on the right In addition, a and c give (5/2)O 2 on the right to cancel out the (5/2)O 2 on the left. 3) Apply all the changes listed above: 3NO 2 (g) > 3NO(g) + (3/2)O 2 (g) N 2 (g) + (5/2)O 2 (g) + H 2 O(l) > 2HNO 3 (aq) 2NO(g) > N 2 (g) + O 2 (g) ΔH = +174 kj ΔH = 128 kj ΔH = 183 kj 4) Add the equations and the ΔH values: +174 + ( 128) + ( 183) = 137 kj 3NO 2 (g) + H 2 O(l) > 2HNO 3 (aq) + NO(g) ΔH = 137 kj Example #5: Calculate ΔH for this reaction: CH 4 (g) + NH 3 (g) > HCN(g) + 3H 2 (g) given: N 2 (g) + 3 H 2 (g) > 2 NH 3 (g) C(s) + 2 H 2 (g) > CH 4 (g) H 2 (g) + 2 C(s) + N 2 (g) > 2 HCN(g) ΔH = 91.8 kj ΔH = 74.9 kj ΔH = +270.3 kj 1) Analyze what must happen to each equation: a) first eq flip and divide by 2 (puts one NH 3 on the reactant side) b) second eq flip (puts one CH 4 on the reactant side) c) third eq divide by 2 (puts one HCN on the product side) 2) rewite all equations with the changes: NH 3 (g) > (1/2)N 2 (g) + (3/2)H 2 (g) CH 4 (g) > C(s) + 2 H 2 (g) ΔH = +45.9 kj < note sign change & divide by 2 ΔH = +74.9 kj < note sign change
(1/2)H 2 (g) + C(s) + (1/2)N 2 (g) > HCN(g) ΔH = +135.15 kj < note divided by 2 3) What cancels when you add the equations: (1/2)N 2 (g) first and third equations C(s) second and third equations (1/2)H 2 (g) on the left side of the third equation cancels out (1/2)H 2 (g) on the right, leaving a total of 3H 2 (g) on the right (which is what we want) 4) Calculate the ΔH for our reaction: +45.9 kj plus +74.9 kj plus +135.15 = 255.95 kj = 260. kj (note three sig figs)
1. What is the increase in entropy of one gram of ice at O o C is melted and heated to 50 0 C? The change in entropy is given by ds = dq T. In this case, the dq must be calculated in two pieces. First there is the heat needed to melt the ice, and then there is the heat needed to raise the temperature of the system. Therefore, Q = ml f + mc w T where m = 1 g is the total mass of the system, L f = 80 cal/g is the heat of fusion, and c w = 1 cal/(g K) is the specific heat of water. Thus, the total change in entropy is S = ml Tf f dt + mc w T i T i T where T i = 0 o C = 273K and T f = 50 o C = 323K. Plugging in numbers, and remembering that we always use temperatures in Kelvin, S = (1 g)(80 cal/g) 273K = 80 cal/k + ln 273 S = 0.461 cal/k 323 + 273 ( 323 273 (1 cal/g)(1 g) dt T ) cal/k 2. Find the change in entropy if 500 g of water at 80 o C is added to 300 g of water at 20 o C. The total amount of water is 800g, so the final temperature of the system is given by ( ) ( ) 5 3 353K + 293K = 330.5K 8 8 For m 1 = 500 g and m 2 = 300 g, the entropy change is given by dq S = T = 330.5 353 c w m 1 dt T + 330.5 293 dt c w m 2 T ). = c w ( m 1 ln 330.5 353 + m 2 ln 330.5 293 c w is the specific heat of water which is 1 cal/(g K), so substituting in numbers, S = 3.20 cal/k. (Please note this answer is very sensitive to roundoff. Changing 330.5K to 331K gives 4.4 cal/k.) 3. Consider a mole of a gas initially at 1 (P 1, V 1 ) and finally at 2 (P 2, V 2 ). Since S 2 S 1 is path independent, choose the simple path shown in Figure (1) by first changing pressure at constant volume and then volume at constant pressure. Let 0 = (P 0, T 0 ) be the intermediate point you go through. Show that S 2 S 1 = C P ln ( T2 T 1 ) R ln Show that if 1 and 2 lie on an adiabatic curve, this difference vanishes. Assume C p = C V + R, but not a particular value to C V. ( P2 P 1 ).
We know from the first law of thermodynamics that du = dq P dv. Therefore, dq = du + P dv. For an ideal gas, du = C v NdT and P = NRT/V. For one mole of gas we have that dq = C v dt + RT dv V. Now, ds = dq/t, so dividing the above equation by T we have ds = C v dt T + R dv V. (1) From here, you could plug in that T = P V/(NR) and dt = 1 NR (P dv + V dp ) to get ds = C v dp P + (C v + R) dv V. Then you could first integrate along constant pressure (dp = 0) and then constant volume (dv = 0) and simplify the resulting expression to find the answer. However, there is no reason we cannot directly integrate equation 1 from point 1 to point 2 to find that Now, we know that V2 V 1 = T2P1 T 1P 2 where C p = C v + R.. S 2 S 1 = C v ln T 2 T 1 + R ln V 2 V 1. by the ideal gas law. Plugging this in for V2 V 1 S 2 S 1 = C v ln T 2 T 1 + R ln T 2P 1 T 1 P 2 = C v ln T 2 T 1 + R ln T 2 T 1 + R ln P 1 P 2 S 2 S 1 = C p ln T 2 T 1 R ln P 2 P 1 we have If 1 and 2 are on an adiabatic curve, then dq = 0 which means that we know ds must equal zero. We see this by remembering that then P V γ = constant where γ = C p /C v. Therefore, P 2 P 1 = ( ) γ V1 V 2 T 2 T 1 = P 2V 2 P 1 V 1 = ( V1 V 2 ) γ 1. Plugging this into the equation for S 2 S 1 we find that ( ) γ 1 V1 S 2 S 1 = C p ln R ln as desired. V 2 ( ) γ V1 V 2 = C p (γ 1) ln V 1 V 2 Rγ ln V 1 V 2 = C p ( C p C v C v ) ln V 1 V 2 Rγ ln V 1 V 2 = R( C p C v ) ln V 1 V 2 Rγ ln V 1 V 2 = Rγ ln V 1 Rγ ln V 1 V 2 V 2 S 2 S 1 = 0
P 1 0 2 V FIG. 1: To compute entropy difference S 2 S 1 go from 1 to 0 at constant volume and then from 0 to 2 at constant pressure.
Germain Henri Hess, in 1840, discovered a very useful principle which is named for him: The enthalpy of a given chemical reaction is constant, regardless of the reaction happening in one step or many steps. There is a second way to use Hess' Law and it is usually presented like this: ΔH rxn = Σ ΔH f (products) minus Σ ΔH f (reactants) Here's another form in which you see this form of Hess' Law: ΔH rxn = Σ ΔH f, products minus Σ ΔH f, reactants The "rxn" above is a common way to abbreviate "reaction." All it means is that we are discussing the enthalpy of a generic reaction, not any specific one. Problem #1: Calculate the standard enthalpy of combustion for the following reaction: C 2 H 5 OH (l) + (7/2) O 2 (g) > 2 CO 2 (g) + 3 H 2 O (l) Before launching into the solution, notice I used "standard enthalpy of combustion." This is a very common chemical reaction, to take something and combust (burn) it in oxygen. It is so common that the phrase "standard enthalpy of combustion" is used alot and is given this symbol: ΔH comb. The key to solving this problem is to have a table of standard enthalpies of formation handy. In case you missed it, look at the equation up near the top and see the subscripted f. What we are going to do is sum up all the product enthalpies of formation and then subtract the summed up reactant enthalpies of formation. Also, we need to have the equation balanced, so be sure to remember to check for that. Fractional coefficients are OK. Like this: ΔH comb = [ 2 ( 393.5) + 3 ( 286) ] minus [ ( 278) + (7/2) (0) ] The boldfaced values are the coefficients and the other ones are the standard enthalpy of formation for the four substances involved. Since oxygen is an element in its standard state, its enthalpy of formation is zero. Doing the math gives us ΔH comb = 1367 kj/mol of ethyl alcohol. The ΔH f values ( 393.5, 286, 278 and zero) were looked up in a reference source. Problem #2: Calculate the standard enthalpy of combustion for the following reaction: C 6 H 12 O 6 (s) + 6 O 2 (g) > 6 CO 2 (g) + 6 H 2 O (l) To solve this problem, we must know the following ΔH f values:
C 6 H 12 O 6 (s) 1275.0 O 2 (g) zero CO 2 (g) 393.5 H 2 O (l) 285.8 All the above values have units of kj/mol because these are standard values. All standard enthalpies have the unit kj/mol. As a brief reminder, here is the chemical reaction for the standard enthalpy of glucose: 6C(s, graphite) + 6H 2 (g) + 3O 2 (g) > C 6 H 12 O 6 (s) Each standard enthalpy value is associated with a chemical reaction. The reaction will always form one mole of the target substance (glucose in the example) in its standard state. The target substance is always formed from elements in their respective standard states. Note how the standard state for carbon is graphite, not diamond or buckerministerfullerene. Remember also that all elements in their standard state have an enthalpy of formation equal to zero. The solution ΔH comb = [ 6 ( 393.5) + 6 ( 285.8) ] minus [ ( 1275) + (6) (0) ] The boldfaced values are the coefficients and the other ones are the standard enthalpy of formation for the four substances involved. Since oxygen is an element in its standard state, its enthalpy of formation is zero. Doing the math gives us ΔH comb = 2801 kj/mol of glucose. Problem #3: Calculate the standard enthalpy of formation for glucose, given the following values: ΔH comb, glucose = 2800.8 kj/mol ΔH f, CO2 = 393.5 ΔH f, H2 O = 285.8 2800.8 = [ 6 ( 393.5) + 6 ( 285.8) ] minus [ (ΔH f, glucose ) + (6)(0) ] Did you see what I did. All the enthalpies of formation are on the right hand side and the ΔH comb goes on the left hand side. By the way, this is a common test question. Be prepared. Here's what happened: 1) First of all, this is the reaction we want an answer for: 6C(s, graphite) + 6H 2 (g) + 3O 2 (g) > C 6 H 12 O 6 (s) We know this because the problem asks for the standard enthalpy of formation for glucose. The above
chemical reaction IS the standard formation reaction for glucose. We want the enthalpy for it. 2) Here are the reactions to be added, in the manner of Hess' Law: C 6 H 12 O 6 (s) + 6 O 2 (g) > 6 CO 2 (g) + 6 H 2 O (l) C(s, gr.) + O 2 (g) > CO 2 (g) H 2 (g) + 1/2 O 2 (g) > H 2 O(l) 3) Flip the first reaction and multiply the other two by six. Then add the three reactions together. If you do it right, you should recover the reaction mentioned just above in (1). Problem #4: Complete combustion of 1.00 mol of acetone (C 3 H 6 O) liberates 1790 kj: C 3 H 6 O (l) + 4 O 2 (g) > 3 CO 2 (g) + 3 H 2 O (l); ΔH comb, acetone = 1790 kj Using this information together with the data below (values in kj/mol), calculate the enthalpy of formation of acetone. ΔH f, O2 : 0 ΔH f, CO2 : 393.5 ΔH f, H2 O : 285.83 1) Hess' Law: ΔH rxn = Σ ΔH f, products minus Σ ΔH f, reactants 2) Sustitute values into equation: 1790 = [ 3 ( 393.5) + 3 ( 285.83) ] minus [ (ΔH f, acetone ) + (4)(0) ] 1790 = 2037.99 ΔH f, acetone 247.99 = ΔH f, acetone ΔH f, acetone = 247.99 kj/mol To three sig figs, the value is 248 kj/mol. Problem #5: The standard enthalpy of formation of hexane can be determined indirectly. Calculate the standard enthalpy of formation of hexane using the enthalpies of combustion (in kj/mol) given just below. C 6 H 14 (l) 4163.0 C(s, gr.) 393.5 H 2 (g) 285.8 Before the solution is given, a bit of discussion: the enthalpy of combustion for hexane, carbon and
hydrogen are these chemical equations: C 6 H 14 (l) + (19/2)O 2 (g) > 6CO 2 (g) + 7H 2 O(l) C(s, gr) + O 2 (g) > CO 2 (g) H 2 (g) + (1/2)O 2 (g) > H 2 O(l) To obtain the target reaction (see just below, in the solution), we must do the following: a) reverse the first equation b) multiply the second equation by 6 c) multiply the third equation by 7 By the way, the second equation (presented as the enthalpy of combustion of carbon) is also the equation for the formation of carbon dioxide. The third equation (presented as the combustion of hydrogen gas) is also the formation equation for water in its standard state (liquid). The moral of the story? Sometimes terms overlap. The 393.5 value is the enthalpy for the combustion of carbon. It is also the formation equation for carbon dioxide. Last point: notice how the enthalpy of combustion focuses on the reactant while the standard enthalpy of formation focuses on the product. 1) Write the equation for the formation of hexane: 6C(s) + 7H 2 (g) > C 6 H 14 (l) Use Hess' Law: ΔH rxn = Σ ΔH comb, products minus Σ ΔH comb, reactants ΔH rxn = [ (4136) ] minus [ (6) ( 393.5) + (7) ( 285.8) ] ΔH rxn = 198.6 kj/mol Problem #6: Ammonia reacts with oxygen to form nitrogen dioxide and steam, as follows: 4 NH 3 (g) + 7 O 2 (g) > 4 NO 2 (g) + 6H 2 O(g) Given the following standard enthalpies of formation (given in kj/mol): NH 3 (g) 45.90 NO 2 (g) +33.1 H 2 O(l) 241.83 calculate the enthalpy of the reaction. Note that water is given as a gas. Normally, water as a liquid is used in problems. Not in this one.
Use Hess' Law: ΔH rxn = [ (4) (+33.1) + (6) ( 241.83) ] minus [ (4) ( 45.90) + (7)(0) ] ΔH rxn = 1134.98 kj = 1135 kj Note that the units kj/mol are NOT used. Problem #7: The standard enthalpy change, ΔH, for the thermal decomposition of silver nitrate according to the following equation is +78.67 kj: AgNO 3 (s) > AgNO 2 (s) + (1/2)O 2 (g) The standard enthalpy of formation of AgNO 3 (s) is 123.02 kj/mol. Calculate the standard enthalpy of formation of AgNO 2 (s) 1) Let's write what we know: AgNO 3 (s) > AgNO 2 (s) + (1/2)O 2 (g) ΔH = +78.67 kj Ag(s) + (1/2)N 2 (g) + (3/2)O 2 (g) > AgNO 3 (s) ΔH f = 123.02 kj 2) Let's write the formation equation for AgNO 2 (s): Ag(s) + (1/2)N 2 (g) + O 2 (g) > AgNO 2 (s) ΔH f =??? 3) Determine the unknown value by adding the two equations listed in step 1: +78.67 kj + ( 123.02 kj) = 44.35 kj (this is the answer) When the two equations are added together, the AgNO 3 (s) cancels out as does (1/2)O 2 (g) and we are left with the formation equation for AgNO 2 (s), the equation given in step 2. Problem #8: Using standard enthalpies of formation, calculate the heat of combustion per mole of gaseous water formed during the complete combustion of ethane gas. The enthalpies of formation needed are: C 2 H 6 (g) 84.68 O 2 (g) zero CO 2 (g) 393.5 H 2 O (g) 241.8 1) The balanced equation for the combustion of C 2 H 6 (ethane) is:
2C 2 H 6 + 7O 2 > 4CO 2 + 6H 2 O 2) The enthalpy of the reaction is: [sum of enthalpies of formation of products] minus [sum of enthalpies of formation of reactants] [(2 moles CO 2 )( 393.5 kj/mole) + (6 moles H 2 O)( 241.8 kj/mole)] [(2 moles C 2 H 6 ) ( 84.68 kj/mole) + (7 moles O 2 )(0 kj/mole)] 2238 kj ( 169 kj) = 2069 kj 3) However, that's the heat produced when we make 6 moles of H 2 O(g). Therefore, 2069 kj / 6 moles H 2 O = 345 kj / mole H 2 O Problem #9: The ΔH for the following reaction equals 89 kj: IF 7 + I 2 > IF 5 + 2IF In addition, these two standard enthalpies of formation are known: IF 7 = 941 kj IF 5 = 840 kj Determine the ΔH f for IF. Solution #1: 1) The enthalpy of the reaction is: [sum of enthalpies of formation of products] minus [sum of enthalpies of formation of reactants] 2) Inserting values into the above, we find: 89 = [( 840)(1) + (2x)] [( 941)(1) + (0)(1)] 89 = 101 + 2x 2x = 190 x = 95 kj Solution #2: 1) Here are all three data reactions written out in equation form: 1/2 I 2 + 7/2 F 2 > IF 7 ΔH f = 941 kj 1/2 I 2 + 5/2 F 2 > IF 5 ΔH f = 840 kj IF 7 + I 2 > IF 5 + 2IF ΔH = 89 kj and here is the target equation:
1/2 I 2 + 1/2 F 2 > IF ΔH f =? 2) What we need to do is add the three data equations together in such a way as to recover the target equation: a) leave equation 1 untouched b) flip eqation 2 c) leave equation 3 untouched. 3) The result of the above is this: I 2 + F 2 > 2IF and ΔH = 941 + (+840) + ( 89) = 190 kj 4) However, this is not the enthalpy of formation, since that value is always for one mole of the product. This is the answer: ΔH f = 190 / 2 = 95 kj Problem #10: What is the enthalpy change for the following reaction? SiCl 4 (l) + 2H 2 (g) > Si(s) + 4HCl(g) Use the following standard enthalpies of formation: SiCl 4 (l); 687 kj mol 1 HCl(g); 92 kj mol 1 ΔH = [0 + 4( 92)] [ 687 + 2(0)] The zeros are the enthalpies for H 2 and Si. These are elements in their standard sate and in that case, the enthalpy of formaton is always zero. ΔH = +319 kj Bonus Problem: Given the following information: ΔH f kj/mol ΔH f kj/mol Li 2 O(s) 597.9 Li + (aq) 278.5 Na 2 O(s) 416 Na + (aq) 240 K 2 O(s) 361 K + (aq) 251 CO(g) 110.5 CO 2 (g) 393.5 H 2 O(l) 286 OH (aq) 230
1/1/2016 ChemTeam: Hess' Law using standard enthalpies of formation CCl 4 (l) 135 SiO 2 (s) 911 LiCl(s) 409 NaCl(s) 411 KCl(s) 436 Cl (aq) 167 Calculate ΔH for the following reaction: 2Li(s) + 2H 2 O(l) > 2LiOH(aq) + H 2 (g) 1) The key is to see the meaning of 2LiOH(aq): 2LiOH(aq) > 2Li + (aq) + 2OH (aq) 2) That means that, in reality, we want the ΔH for this reaction: 2Li(s) + 2H 2 O(l) > 2Li + (aq) + 2OH (aq) + H 2 (g) 3) We need the following formation reactions: Li(s) > Li + (aq) + e ΔH f = 278.5 kj/mol e + 1/2H 2 (g) + 1/2O 2 (g) > OH (aq) ΔH f = 230 kj/mol 4) Rewrite the revised target equation: 2Li(s) + 2H 2 O(l) > 2Li + (aq) + 2OH (aq) + H 2 (g) 5) Use Hess' Law utilizing the revised target equation: ΔH = [(2) ( 278.5) + (2) ( 230) + (0)] [(2) (0) + (2) ( 286)] ΔH = 445kJ
Problem #1: Suppose a piece of iron with a mass of 21.5 g at a temp of 100.0 C is dropped into an insulated container of water. The mass of the water is 132.0 g and its temperature before adding the iron is 20.0 C. What will be the final temp of the system? Specific heat of iron is 0.449 kj/kg K. 1) Since we write q lost, metal = q gained, water (mass) (Δt) (C p, metal ) = (mass) (Δt) (C p, water ) 2) Substituting: (21.5) (100 x) (0.449) = (132.0) (x 20) (4.184) Some explanation: a) 100 x is the Δt for the metal; it starts at 100.0 C and drops to some unknown, final value. b) x 20 is the Δt for the water; it starts at 20.0 C and rises to some unknown, final value. c) Since both metal and water wind up at the same ending value, we need to use only one unknown for the two Δt expressions. 3) A wee bit of algebra: (2150 21.5x) (0.449) = (132x 2640) (4.184) 965.35 9.6535x = 552.288x 11045.76 561.9415x = 12011.11 To 3 sig figs, the answer is 21.4 C. Problem #2: A 12.48 g sample of an unknown metal, heated to 99.0 C was then plunged into 50.0 ml of 25.0 C water. The temperature of the water rose to 28.1 C. Assuming no loss of energy to the surroundings: 1. How many joules of energy did the water absorb? 2. How many joules of energy did the metal lose? 3. What is the heat capacity of the metal? 4. What is the specific heat of metal? 1) q = (50.0 g) (3.1 C) (4.184 J g 1 C 1 ) = 648.52 J 2) 648.52 J 3) 648.52 J / 70.9 C = 9.147 J/ C
4) 9.147 J/ C divided by 12.48 g = 0.733 J g 1 C 1 Comment #1: this question doesn't use the q lost = q gained formulation of other questions. That is because the question is broken up into four parts. Notice that parts (1) and (2) are the equivalent of q lost = q gained and that (4) is the usual answer sought in problems of this type. Comment #2: (3) is a step unnecessary to the solution for (4). It is there so you notice the difference between heat capacity and specific heat capacity. Problem #3: A 43.2 g block of an unknown metal at 89.0 C was dropped into an insulated vesssel containing 43.00 g of ice and 26.00 g of water at 0 C. After the system had reached equilibrium it was determined that 9.15 g of the ice had melted. What is the specific heat of the metal? (The heat of fusion of ice = 334.166 J g 1.) Comment: this variation of the usual suspects (detailed above) does NOT involve a temperature change in the water, only in the metal. Rather, some ice melts and the whole ice water system stays at zero Celsius. Very interesting! 1) Determine heat gained by the ice that melted: 9.15 g times 334.166 J g 1 = 3057.62 J 2) Substitue and solve for the specific heat: q = (mass) (Δt) (C p, metal ) 3057.62 J = (43.2 g) (89.0 C) (x) x = 0.795 J g 1 C 1 Problem #4: A 35.0 g block of metal at 80.0 C is added to a mixture of 100.0 g of water and 15.0 g of ice in an isolated container. All the ice melted and the temperature in the container rose to 10.0 C. What is the specific heat of the metal? 1) Determine heat required to melt the ice: q = (15.0 g) (334.166 J g 1) = 5012.49 J Note that the 100 g of water is not mentioned yet. 2) Determine heat need to raise 115 g of water from 0 to 10.0 C: q = (115 g) (10.0 C) (4.184 J g 1 C 1) = 4811.6 J Note the inclusion of the melted 15 g of ice. Also, notice that the water was at zero C. We know this from the presence of the ice. 3) Determine the specific heat of the metal:
1/1/2016 ChemTeam: How to Determine Specific Heat: Problem 1 10 (5012.49 J + 4811.6 J) = (35.0 g) (70.0 C) (x) x = 4.01 J g 1 C 1 Problem #5: A 500.0 g sample of an element at 153.0 C is dropped into an ice water mixture. 109.5 g of ice melts and there is still an ice water mixture. What is the specific heat of the metal in J/g C? Given the molar heat capacity of the metal is 26.31 J/mol C, what is the atomic weight and identity of the metal? 1) Determine energy needed to melt the ice: (6.02 kj/mol) (109.5 g / 18.015 g/mol) = 36.5912 kj 2) Determine specific heat: 36591.2 J = (500.0 g) (153.0 C) (x) x = 0.4783 J/g C Note: we know the change in temperature is 153.0 C because there is still ice in the water. That means the ice water mix remained at zero Celsius as the 109.5 g of ice melted. 3) Determine aomic weight of the element: 0.4783 J/g C times x = 26.31 J/mol C x = 55.0 g/mol The element is manganese. Problem #6: A 12.48 g sample of an unknown metal is heated to 99.0 C and then was plunged into 50.0 ml of 25.0 C water. The temperature of the water rose to 28.1 C. (a) How many joules of energy did the water absorb? (b) How many joules of energy did the metal lose? (c) What is the heat capacity of the metal? (d) What is the specific heat capacity of the metal? 1) Solution to (a): q = (50.0 g) (3.1 C) (4.181 J g 1 C 1) = 648.52 J I used 50.0 g because the density of water is 1.00 g/ml and I had 50.0 ml of water. 2) Solution to (b): q = 648.52 J We assume all heat absorbed by the water was lost by the metal. We assume no loss of heat energy to the outside during the transfer.
3) Solution to (c): 648.52 J / 74.0 C = 8.76 J/ C (or 8.76 J/K) 4) Solution to (d): (50.0 g) (3.1 C) (4.181 J g 1 C 1) = (12.48 g) (74.0 C) (x) Solve for x. Problem #7: What is the specific heat of a metal if addition of 90.0 g of the metal at 17.7 C to 210.0 g of Cu (s = 0.385 J/g C) at 153.7 C produces a mixture that reaches thermal equilibrium at 129.1 C? Comment: notice that the two metals are being added to each other. Imagine a situation where each sample is composed of dust or very small pellets. Then, the two dry samples are rapidly mixed together. (90.0 g) (111.4 C) (x) = (210.0 g) (24.6 C) (0.385 J/g C) x = 0.198 J/g C Problem #8: A 31.0 gram block of an unknown metal at 88.0 C was dropped into an insulated flask containing approximately 30.0 grams of ice and 20.0 grams of water at 0.0 C. After the system had reached a steady temperature, it was determined that 12.1 grams of ice had melted. What is the specific heat of the metal? The heat of fusion of ice is equal to 334.166 J/g. 12.0 g times 334.166 J/g = 4009.992 J 4009.992 J = (31.0 g) (88.0 C) (x) x = 1.47 J / g C Comment: the fact that ice remained in the water when the temperature reached equilibrium means the water ice mixture stayed at zero Celsius. This means the metal went from 88.0 C to 0 C, for a Δt of 88.0 C Problem #9: A 25.95 g sample of methanol at 35.60 C is added to a 38.65 g sample of ethanol at 24.70 C in a constant pressure calorimeter. If the final temperature of the combined liquids is 28.65 C and the heat capacity of the calorimeter is 19.3 J/C, determine the specific heat of methanol. the heat lost by the methanol goes to (1) heating the ethanol and (2) heating the calorimeter (25.95 g) (6.95 C) (x) = (38.65 g) (3.95 C) (2.44 J g 1 C 1 ) + (3.95 C) (19.3 J/C) x = 2.49 J g 1 C 1