Midterm Exam IV. CHEM 181: Introduction to Chemical Principles November 29, C(s, coal) + O 2 (g) CO 2 (g)

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Clifford Patrick
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1 Midterm Exam IV CHEM 8: Introduction to Chemical Principles November 29, 202. Use some or all of the following data: C(graphite) 0. C(diamond).9 CO 2 (g) H 2 O(g) 24.8 CH 4 (g) 74.6 H f (kj mol ) H bond (kj mol ) C C 347 C H 44 H H 435 (a) Assume H f is 0 for C(s, coal). Calculate the H rxn for the combustion of coal and of CH 4. For the combustion of coal, C(s, coal) + O 2 (g) CO 2 (g) Using heats of formation (and remembering H f is zero for oxygen): = H f (CO 2 (g)) H f (C(s,coal)) = = kj mol For CH 4, CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) and = [ H f (CO 2 (g)) + 2 H f (H 2 O(g)) ] H f (CH 4 (g)) = [ ( 24.8)] [ 74.6] = kj mol

2 (b) Use the data on the previous page to estimate H rxn for Using your answer, calculate 2CH 4 (g) C 2 H 6 (g) + H 2 (g) i. H f for C 2H 6. ii. H rxn for the combustion of C 2 H 6. The reaction breaks two C H bonds and forms one C C and one H H bond. So H rxn = H bond(reactants) H bond(products) = [2 44] [ ] = 46 kj mol Working the previous part backwards (with H f = 0 for H 2): Hrxn = Hf (products) Hf (reactants) 46 kj mol = Hf (C 2 H 6 (g)) 2( 74.6) Hf (C 2 H 6 (g)) = ( 74.6) = 03.2 kj mol (This value is a little more negative than the actual value, because C H bonds in CH 4 are a little stronger than average.) For combustion: and C 2 H 6 (g) O 2(g) 2CO 2 (g) + 3H 2 O(g) = [ 2 H f (CO 2 (g)) + 3 H f (H 2 O(g)) ] H f (C 2 H 6 (g)) = [2( 393.5) + 3( 24.8)] [ 03.2] = 409 kj mol (c) Make a quantitative argument using enthalpies of formation and/or bond enthalpies that supports the statement: 2

3 Combustion of coal produces more CO 2 (g) per unit of energy than any other fossil fuel. Your argument should go beyond the two examples here (CH 4 and C 2 H 6 ). The previous part s approach can be extended to any hydrocarbon: RH + CH 4 (g) RCH 3 + H 2 (g) Given the approximations involved with using bond enthalpies, this has the same +46 kj mol reaction enthalpy. Heat of formation is then: 46 kj mol = H f (RCH 3 ) ( H f (RH) 74.6) H f (RCH 3 ) = H f (RH) 28.6 kj mol So comparing the two reactions, and RH+?O 2 (g)?co 2 (g)+?h 2 O(g) RCH 3 +?O 2 (g)?co 2 (g)+?h 2 O(g) the second one will produce one extra CO 2 and one extra H 2 O that is, the enthalpy of formation of the combustion products will be = kj mol more negative, and the reactants only 28.6 kj mol more negative. That extra CO 2, then, is accompanied by kj mol extra released heat; that s much more than the kj mol per CO 2 released burning coal. Another approach would make fossil-fuel hydrocarbons from coal by first reacting the coal with H 2. For each C C and H H bond broken, there are two new C H bonds formed; this is the reverse of the calculation done in part (b), and so the reaction enthalpy is 46 kj mol per H 2 added. This does not change the amount of CO 2 produced upon combustion, but because it produces an extra water, it makes the reaction enthalpy more negative by = 95.8 kj mol. 2. A L sample of M Ba(NO 3 ) 2 is added to L of M MgSO 4 in a calorimeter with a total heat capacity of 455 J K. The observed increase in temperature is.43 K. 3

4 (a) Calculate the value of H rxn (ignoring any dependence on concentration) for the equation Ba(NO 3 ) 2 (aq) + MgSO 4 (aq) BaSO 4 (s) + Mg(NO 3 ) 2 (aq) (You will need this value of H rxn for the next part of the problem. If you do not get an answer, you can continue the calculation assuming H rxn = 30 kj mol.) There are ( (5 0 2 L) 0.5 mol ) = mol L of BaSO 4 (s) formed. The total heat released is q = C T = (455 J K )(.43 K) = J Since the surroundings are increasing in temperature, enthalpy change for the system is negative, and J H rxn = mol = 26 kj mol (b) At equilibrium at 20 C, the concentration of a saturated solution of BaSO 4 in water is M. What is the concentration of a saturated solution at 80 C? Change these temperatures to Kelvin, right off: 293 K and 353 K. The important equilibrium is Ba 2+ (aq) + SO 2 4 (aq) BaSO 4 (s) with K = [Ba 2+ ][SO 2 4 ] since we know the equilibrium concentrations at 20 C, we know that K 293K = = ( ) 2 4

5 This is very large, in keeping with a sparingly soluble BaSO 4. Since the reaction as written releases heat, LeChatelier s principle says that raising temperature should push equilibrium towards reactants that is, greater solubility. Calculating the temperature dependence of equilibrium requires the van t Hoff equation: ln K 353K K 293K = H R ( 353 K ) 293 K = J mol ( J mol K K ) =.84 K 353K K 293K = e.84 = 0.63 K 353K = 0.63K 293K = Plugging this back into the definition of K gives [Ba 2+ ] = [SO 2 4 ] = = M (It might be conceptually easier to reverse the reaction deal with solvation rather than precipitation and flip the sign of H.) 3. The equilibrium constant at 25.0 C for the equation Co 3+ (aq) + 6NH 3 (aq) [Co(NH 3 ) 6 ] 3+ (aq) is K f = M 6 (a) Calculate the value of G rxn at 25.0 C. In which direction is the reaction spontaneous when Co 3+ (aq), NH 3 (aq), and [Co(NH 3 ) 6 ] 3+ are at standard conditions? G rxn = RT ln K = (8.34 J K )(298.5 K) ln( ) = 4.67 kj mol 5

6 Reaction is spontaneous in the forward direction, as written. (b) Calculate the value of G rxn when [Co 3+ ] = M, [NH 3 ] = 0. M, and [[Co(NH 3 ) 6 ] 3+ ] =.00 M. In what direction is the reaction spontaneous under these conditions? Q = [Co(NH 3) 3+ 6 ] [Co 3+ ][NH 3 ] 6 = = G rxn = RT ln Q K = (8.34 J K )(298.5 K) ln = 5.7 kj mol Reaction is spontaneous in the backward direction (products to reactants), as written. 6

Enthalpies of Reaction

Enthalpies of Reaction Enthalpy is an extensive property Magnitude of H is directly related to the amount of reactant used up in a process. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) H = 890 kj 2CH 4 (g)