AP Practice Questions

Transcription

1 1) AP Practice Questions The tables above contain information for determining thermodynamic properties of the reaction below. C 2 H 5 Cl(g) + Cl 2 (g) <===> C 2 H 4 Cl 2 (g) + HCl(g) (a) Calculate ΔH for the reaction above, using the table of average bond dissociation energies. (b) Calculate ΔS for the reaction at 298 K, using data from either table as needed. (c) Calculate the value for K eq for the reaction at 298 K. (d) What is the effect of an increase in temperature on the value of the equilibrium constant? Explain your answer. 2) BCl 3 (g) + NH 3 (g) <---> Cl 3 BNH 3 (s) The reaction represented above is a reversible reaction. (a) Predict the sign of the entropy change, ΔS, as the reaction proceeds to the right. Explain your prediction. (b) If the reaction is thermodynamically favored to proceed to the right, predict the sign of the enthalpy change, ΔH. Explain your prediction. (c) The direction in which the reaction is thermodynamically favored to proceed changes as the temperature is increased above a specific temperature. Explain. (d) What is the value of the equilibrium constant at the temperature referred to in (c); that is, the specific temperature at which the thermodynamically favored direction of the reaction changes? Explain. 3) Cl 2 (g) + 3 F 2 (g) ---> 2 ClF 3 (g) ClF 3 can be prepared by the reaction represented by the equation above. For ClF 3 the standard enthalpy of formation, ΔH f, is kilojoules/mole and the standard free energy of formation, ΔG f, is kilojoules/mole. (a) Calculate the value of the equilibrium constant for the reaction at 298 K. (b) Calculate the standard entropy change, ΔS, for the reaction at 298 K. (c) If ClF 3 were produced as a liquid rather than as a gas, how would the sign and magnitude of ΔS for the reaction be affected? Explain. (d) at 298 K the absolute entropies of Cl 2 (g) and ClF 3 (g) are joules per mole-kelvin and joules per mole-kelvin, respectively. (i) Account for the larger entropy of ClF 3 (g) relative to that of Cl 2 (g). (ii) Calculate the value of the absolute entropy of F 2 (g) at 298 K.

2 4) Br 2 (l) ---> Br 2 (g) At 25 C the equilibrium constant, K p, for the reaction above is atmosphere. (a) What is ΔG 298 for this reaction? (b) It takes 193 joules to vaporize 1.00 gram of Br 2 (l) at 25 C and 1.00 atmosphere pressure. What are the values of ΔH 298 and of ΔS 298 for this reaction? (c) Calculate the normal boiling point of bromine. Assume that ΔH and ΔS remain constant as the temperature is changed. (d) What is the equilibrium vapor pressure of bromine at 25 C? 5) Substance Enthalpy of Combustion ΔH Kilojoules/mole Absolute Entropy, S Joules/mole-K C(s) H 2 (g) C 2 H 5 OH(l) H 2 O(l) (a) Write a separate, balanced chemical equation for the combustion of each of the following: C(s), H 2 (g), and C 2 H 5 OH(l). Consider the only products to be CO 2 (g) and/or H 2 O(l). (b) In principle, ethanol can be prepared by the following reaction. 2 C(s) + 2 H 2 (g) + H 2 O(l) --> C 2 H 5 OH(l) Calculate the standard enthalpy change, ΔH, for the preparation of ethanol, as shown in the reaction above. (c) Calculate the standard entropy change, ΔS, for the reaction given in part (b) (d) Calculate the value of the equilibrium constant at 25 C for the reaction represented by the equation in part (b). 6) CO(g) + 2 H 2 (g) <===> CH 3 OH(l) For this reaction, ΔH = kilojoules ΔH f (kj mol 1) ΔG f (kj mol 1) S (J mol 1 K 1) CO(g) CH 3 OH(l) The data in the table above were determined at 25 C. (a) Calculate ΔG for the reaction above at 25 C. (b) Calculate K eq for the reaction above at 25 C. (c) Calculate ΔS for the reaction above at 25 C. (d) In the table above, there are no data for H 2. What are the values of ΔH f, ΔG f, and of the absolute entropy, S, for H 2 at 25 C?

3 7) Substance Standard Heat of Formation, ΔH f, in kj mol 1 Absolute Entropy S, in J mol 1 K 1 C(s) CO 2 (g) H 2 (g) H 2 O(l) O 2 (g) C 3 H 7 COOH(l)? The enthalpy change for the combustion of butyric acid at 25 C, ΔH comb, is -2,183.5 kilojoules per mole. The combustion reaction is C 3 H 7 COOH(l) + 5 O 2 (g) <===> 4 CO 2 (g) + 4 H 2 O(l) (a) From the data above, calculate the standard heat of formation, ΔH f, for butyric acid at 25 C. (b) Write a correctly balanced equation for the formation of butyric acid from its elements. (c) Calculate the standard entropy change, ΔS f, for the formation of butyric acid at 25 C. The entropy change, ΔS, for the combustion reaction above is J K 1 at 25 C. (d) Calculate the standard free energy of formation, ΔG f, for butyric acid at 25 C. 8) 2 H 2 S(g) + SO 2 (g) <===> 3 S(s) + 2 H 2 O(g) At 298 K, the standard enthalpy change, ΔH, for the reaction represented above is -145 kilojoules. (a) Predict the sign of the standard entropy change, ΔS, for the reaction. Explain the basis for your prediction. (b) At 298 K, the forward reaction (i.e., toward the right) is thermodynamically favored. What change, if any, would occur in the value of ΔG for this reaction as the temperature is increased? Explain your reasoning using thermodynamic principles. (c) What change, if any, would occur in the value of the equilibrium constant, K eq, for the situation described in (b)? Explain your reasoning. (d) The absolute temperature at which the forward reaction is no longer thermodynamically favored can be predicted. Write the equation that is used to make the prediction. Why does this equation predict only an approximate value for the temperature?

4 1) a) two points [delta]h = bonds broken minus bonds formed. C 2 H 5 Cl + Cl-Cl ---> C 2 H 4 Cl 2 + HCl [delta]h = ( ) minus ( ) [delta]h = = kj mol 1 CH + Cl-Cl ---> C-Cl + HCl [delta]h = ( ) minus ( ) [delta]h = kj mol 1 b) four points [delta]g = [[delta]g f C 2 H 4 Cl 2 + [delta]g f HCl] - [[delta]g f C 2 H 5 Cl + [delta]g f Cl 2 ] = ( ) - ( ) = kj [delta]g = [delta]h - T[delta]S [delta]s = ((- 151 kj - (- 115 kj)) / 298 = kj mol 1 K 1 [delta]g = - RT ln K - ln K = / (8.314 x 298) ln K = K = 1.50 x d) one point K eq will decrease with an increase in T because the reverse (endothermic) reaction will be favored with addition of heat [delta]g will be less negative with an increase in temperature (from [delta]g = [delta]h - T[delta]S) which will cause K to decrease. 2) a) two points ΔS will be negative. The system becomes more ordered as two gases form a solid. b) two points ΔH must be negative. For the reaction to be thermodynamically favored, ΔG must be negative, so ΔH must be more negative than -TΔS is positive. As T increases, -TΔS increases. Since ΔS is negative, the positive -TΔS term will eventually exceed ΔH (which

5 is negative), making ΔG positive. (In the absence of this, ΔG = ΔH - TΔS and general discussion of the effect of T and ΔS gets 1 point.) d) two points The equilibrium constant is 1. The system is at equilibrium at this temperature with an equal tendancy to go in either direction. ΔG = 0 at equilibrium so K = 1 in ΔG = -RT ln K (In the absence of these, ΔG = -RT ln K gets 1 point). The above concludes the AP scoring standards published in The following is simply alternate ways of answering which the AP readers may or may not have given full credit to. a) The amount of entropy goes down, ΔS is negative. b) ΔG = ΔH - TΔS. If ΔS is negative, then ΔH must also be negative to get a negative ΔG. c) Let us say ΔG is positive when ΔH is positive and ΔS is positive. As T goes up - TΔS becomes more negative until it makes ΔG (which equals ΔH - TΔS) become negative. d) At the temperature when the direction changes, the rate forward = the rate reverse. Since K = k f / k r, this equals 1. 3) average = 3.5 (Only 30 scores of nine; kids did not see stoichiometry in (b), had problems on which gas constant to use, and a hard time in (c) in relating a more negative value.) a) two points [delta]g = - RT ln K; rearranging gives ln K = [delta]g - RT ln K = - 246,000 J - ((8.31 J/mol K) (298 K)) = 1.32 x b) two points [delta]g = [delta]h - T[delta]S - 246,000 J = - 326,400 J - (298) (x) x = J / K [delta]s is a larger negative number ClF 3 (liquid) is more ordered (less disordered) than ClF 3 (gas) This was my answer before I saw the standard given above: A liquid is more ordered than a gas. There would be a greater entropy change in gas + gas ---> liquid than in gas + gas ---> gas. Therefore, sign is the same, but absolute magnitude is greater. d) three points i) ClF 3 is a more complex molecule (i.e. more atoms) with more vibrational and rotational degrees of freedom than Cl 2 ii) Cl F 2 ---> 2 ClF 3 ; use Hess's Law [sigma]s rxn = [sigma]s products - [sigma]s reactants = [2 (281.50)] - [ x] x = 203 J mol 1 K 1

6 4) Average score = 3.2 a) two points [delta]g = - RT lnk = - (8.31 J mol 1 K 1) (298 K) (ln 0.281) = 3.14 x 10 3 J / mol b) four points [delta]h = (193 J/g) (160. g/mol) = 3.08 x 10 4 J / mol [delta]g = [delta]h - T [delta]s [delta]s = ([delta]h - [delta]g ) / T = [(3.084 x 10 4 ) - (3.14 x 10 3 )] / 298 = (2.770 x 10 4 J / mol) / 298 K = 92.9 J / mol K At boiling pt, [delta]g = 0 and thus, T = [delta]h / [delta]s = 3.08 x 10 4 / 92.9 = 332 K d) one point Vapor pressure = atm Note added for student's benefit: this comes directly from the fact that K p = P Br2. 5) Average score = 3.25 a) two points C + O 2 --> CO 2 2 H 2 + O 2 --> 2 H 2 O C 2 H 5 OH --> 2CO H 2 O (Half credit for recognition that combustion is reaction with O 2 ) b) three points 2 C + 2 O 2 ---> 2 CO 2 [delta]h = 2 ( ) = kj 2 H 2 + O 2 ---> 2 H 2 O [delta]h = 2( ) = kj 2 CO O 2 ---> C 2 H 5 OH + 3 O 2 [delta]h = - ( ) = kj Sum of three equations above 2 C + 2 H 2 + H 2 O ---> C 2 H 5 OH [delta]h = kj (1 point for correct [delta]h for each of the first 3 reations) [delta]h combustion C(s) = [delta]h f CO 2 (g)

7 [delta]h combustion H 2 (g) = [delta]h f H 2 O(l) C 2 H 5 OH + 3 O 2 ---> 2 CO H 2 O [delta]h = -1,366.7kJ [delta]h reaction = [sigma] [delta]h f (products) - [sigma] [delta]h f (reactants) = [2( ) + 3( ) - [delta]h f C2H5OH - 0] kj = kj 2 C + 2 H 2 + H 2 O ---> C 2 H 5 OH [delta]h reaction = [sigma] [delta]h (products) - [sigma] [delta]h (reactants) = [ ( )] kj = kj c) one point [delta]s = [sigma] S (products) - [sigma] S (reactants) = ( ) J / mol K = J / mol K d) three points (1) [delta]g = [delta]h - T [delta]s = 8,100 J - (298) ( ) J = 8,100 J + 54,200 J = 62,300 J (2) [delta]g = - RTlnK lnk = - [delta]g / RT = J [(8.31J/mol-K) (298K)] = K = 1.1 x [delta]g = RT logk logk = - [delta]g / RT = J [(2.303) (8.31) (298)] = K = 1.3 x If the terms are not rounded, K = 1.2 x correct substitutions in (1) and (2) earns one point 6) a) two points [delta]g = [sigma] [delta]gf (prod) - [sigma] [delta]gf (react) [delta]g = ( (0)) [delta]g = KJ/mol b) two points [delta]g = - RT ln K (or RT log K) = - ( 8.31 x 10 3) (298) ln K ln K = K = 1.17 x 10 5

8 [delta]g = [delta]h - T [delta]s - 28,900 = -128, [delta]s [delta]s = - 99,200 / 298 = J / mol-k d) three points [delta]hf (H2) = 0 and [delta]gf (H2) = 0 deltas = [sigma]s (prod) - [sigma]s (react) J / mol-k = J / mol-k J / mol-k - 2 S (H2) S = 131 J / mol-k 7) Average score 5.52 a) three points From Hess's law: [delta]h f = [4 (393.5) + 4 (205.85)] kj = kj b) one point 4 C(s) + 4 H 2 (g) + O 2 (g) ---> C 3 H 7 COOH [delta]s f butyric acid = S butyric acid - 4 S carbon - 4 S H 2 - S O 2 [delta]s f butyric acid = [ (5.69) - 4(130.6) - 205] joules/k = joules/k Note: If part c was based on the equation in part b, and was done correctly, credit was given for part c even if part b was wrong. d) three points [delta]g f butyric acid = [delta]h f - T [delta]s f [delta]g f = [ (298) ( )] kj [delta]g f = kj Note: For each of the problems, a maximum of one point was subtracted for gross misuse of significant figures and. or for a mathematical error if correct principles were used. 8) Note: for parts (a), (b), and (c), just writing an equation is not sufficient for the 'explanation" point. To earn credit, the student must connect the equation to the issue to be explained. a) two points Statement that [delta]s is negative 3 moles of gas ---> 2 moles of gas plus solid (3 moles ---> 2 moles earns no points), 2 gases ---> one gas + solid use of [delta]g = [delta]h - T[delta]S with [delta]g = 0 b) two points [delta]g is less negative, goes to 0, goes positive, gets larger Explanation using [delta]g = [delta]h - T[delta]S

9 K eq decreases (exponent goes more negative) as T increases K eq goes from > 1, to 1, to < 1, as T increases Correct explanation using the equation [delta]g = - RT ln K (or ln(k 1 / k 2 ) = ([delta]h / R) (1/T 2-1/T 1 ) higher T favors the reverse reaction (Le Châtelier) because the forward reaction is exothermic. Note: if answer for (a) is that [delta]s is positive, then statement that K eq will decrease or increase depending on the relative magnitudes of T and [delta]g change earns two points. d) two points Since [delta]g = 0 at this point, the equation is T= [delta]h / [delta]s ([delta]g = [delta]h - T [delta]s S is NOT sufficient without [delta]g = 0.) Prediction is not exact since [delta]h and/or [delta]s vary with T.