{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# exam --1 - pokharel(yp624 – Homework 2 – Sutcliﬀe...

This preview shows pages 1–3. Sign up to view the full content.

pokharel (yp624) – Homework 2 – Sutcliffe – (52410) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This HW is due at 11pm the day before Exam 1. 001 10.0 points Calculate the standard Gibbs free energy at 298 K for the reaction 4 NH 3 (g) + 7 O 2 (g) 4 NO 2 (g) + 6 H 2 O( ) Species Δ H 0 f S 0 kJ/mol J/mol · K NH 3 (g) - 46 . 11 192 . 3 O 2 (g) 0 . 0 205 . 0 NO 2 (g) +33 . 2 240 . 0 H 2 O( ) - 285 . 8 69 . 91 1. +1398 kJ / mol rxn 2. - 1152 kJ / mol rxn correct 3. +244 , 363 kJ / mol rxn 4. - 180 , 455 kJ / mol rxn 5. +1152 kJ / mol rxn 6. - 1643 kJ / mol rxn 7. - 825 kJ / mol rxn 8. +1643 kJ / mol rxn 9. - 1398 kJ / mol rxn 10. +825 kJ / mol rxn Explanation: Δ H 0 rxn = summationdisplay n Δ H 0 f prod - summationdisplay n Δ H 0 f rct = bracketleftBig 6( - 285 . 8 kJ / mol) + 4(33 . 2 kJ / mol) bracketrightBig - 4( - 46 . 11 kJ / mol) = - 1397 . 56 kJ / mol Δ S 0 rxn = summationdisplay n Δ S 0 f prod - summationdisplay n Δ S 0 f rct = bracketleftBig 6(69 . 91 J / mol · K) + 4(240 . 0 J / mol · K) bracketrightBig - bracketleftBig 7(205 . 0 J / mol · K) + 4(192 . 3 J / mol · K) bracketrightBig = - 824 . 74 J / mol · K = - 0 . 82474 kJ / mol · K Δ G 0 = Δ H 0 - T Δ S 0 = - 1397 . 56 kJ / mol - (298 K) ( - 0 . 82474 kJ / mol · K) = - 1151 . 79 kJ / mol 002 10.0 points For a given reaction at 300 K, Δ G = - 412 kJ/mol rxn and at 400 K, Δ G = - 439 kJ/mol rxn. The enthalpy for the reaction is the same at both temperatures (Δ H = - 331 kJ/mol rxn). What is the Δ S (entropy) for this reaction? 1. - 27000 J/mol K 2. 27000 J/mol K 3. 270 J/mol K correct 4. 80000 J/mol K 5. 0.0 J/mol K Explanation: Δ G 1 = - 412 kJ / mol rxn T 1 = 300 K Δ G 2 = - 439 kJ / mol rxn T 2 = 400 K Δ G = Δ H - T Δ S Δ S = Δ G - Δ H - T and Δ H = Δ G + T Δ S

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
pokharel (yp624) – Homework 2 – Sutcliffe – (52410) 2 Δ H 1 = Δ H 2 = - 331 kJ / mol rxn Δ S 1 = ( - 412 + 331) kJ / mol rxn - 300 K = 0 . 27 kJ / K · mol rxn = 270 J / mol · K Δ S 2 = ( - 439 + 331) kJ / mol rxn - 400 K = 0 . 27 kJ / K · mol rxn = 270 J / mol · K Alternate Explanation : Δ G 1 + T 1 Δ S = Δ G 2 + T 2 Δ S Δ G 1 - Δ G 2 = T 2 Δ S - T 1 Δ S = Δ S ( T 2 - T 1 ) Δ S = Δ G 1 - Δ G 2 T 2 - T 1 = ( - 412 + 439) kJ / mol rxn (400 - 300) K = 0 . 27 kJ / K · mol rxn = 270 J / mol · K 003 10.0 points For the reaction 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) Δ H r = +198 kJ · mol 1 at 298 K. Which state- ment is true for this reaction? 1. Δ G r will be negative at high tempera- tures. correct 2. The reaction will not be spontaneous at high temperatures. 3. The reaction is driven by the enthalpy. 4. Δ G r will be positive at high tempera- tures. 5. The reaction will not be spontaneous at any temperature. Explanation: Δ G = Δ H - T Δ S is used to predict spon- taneity. G is negative for a spontaneous reaction.) Δ H is positive and T is always pos- itive. For the reaction 2 mol gas 3 mol gas. The more moles of gas, the higher the disor- der, so Δ S is positive and Δ G = (+) - T (+). For Δ G to be negative, T must be large. 004 10.0 points For the reaction 2 C(s) + 2 H 2 (g) C 2 H 4 (g) Δ H r = +52 . 3 kJ · mol 1 and Δ S r = - 53 . 07 J · K 1 · mol 1 at 298 K. The reverse reaction will be spontaneous at 1. temperatures below 1015 K.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

exam --1 - pokharel(yp624 – Homework 2 – Sutcliﬀe...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online