Part A Given the following equation, N 2 O(g) + NO 2 (g) 3 NO(g) G° rxn = -23.0

Question

Part A

Given the following equation,

N2O(g) + NO2(g) 3 NO(g) G°rxn = -23.0 kJ

Calculate G°rxn for the following reaction.

18 NO(g) 6 N2O(g) + 6 NO2(g)

Answer Choices To Choose From: A) -3.83 kJ B) -138 kJ C) -23.0 kJ D) 23.0 kJ E) 138 kJ

Part B

CO2(g)C(s)+O2(g)Hrxn= +393.5 kJ

(Ssys<0, Ssurr<0)

In addition, predict under what temperatures (all temperatures, low temperatures, or high temperatures), if any, the reaction in part D will be spontaneous

A) The reaction is spontaneous at all temperatures.

C) The reaction is spontaneous at high temperatures.

D) The reaction is nonspontaneous at all temperatures.

Explanation / Answer

part A :  answer :  E) 138 kJ

N2O(g) + NO2(g) 3 NO(g) G°rxn = -23.0 kJ

18 NO(g) 6 N2O(g) + 6 NO2(g)

this equation is reverse of the first reaction and multiplied by 6.

so G°rxn = 6 x 23 = 138 kJ

Part B :

answer : D) The reaction is nonspontaneous at all temperatures.

if Hrxn is positive and Srxn negative .

then Grxn = Hrxn - T Srxn

the reaction never be spontenous

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