Enthalpy Q (???)

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unsung

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EK 1001Qs #369:

Under standard conditions, which of the following reactions will have the greatest difference between enthalpy and energy of reaction?

A. 2C(graphite) + O2(g) --> 2CO(g)
B. C(graphite) + O2(g) --> CO2(g)
C. C(graphite) --> C(diamond)
D. CO(g) + NO2(g) --> CO2(g) + NO(g)

The correct answer is A. I don't even get how to tackle the Q, frankly. Why must gen chem be 1000Xs worse than o-chem? help?
 
EK 1001Qs #369:

Under standard conditions, which of the following reactions will have the greatest difference between enthalpy and energy of reaction?

A. 2C(graphite) + O2(g) --> 2CO(g)
B. C(graphite) + O2(g) --> CO2(g)
C. C(graphite) --> C(diamond)
D. CO(g) + NO2(g) --> CO2(g) + NO(g)

The correct answer is A. I don't even get how to tackle the Q, frankly. Why must gen chem be 1000Xs worse than o-chem? help?
I'm going to make an attempt at this question but not sure if this is right:

So Gibbs Free Energy equation says that delta G = delta H - TdeltaS?
In the first equation, 3 moles of material are combining to form product, which means that the change in entropy is negative (or not favored). By rearranging G+TS = H, the entropy term influences the net the energy of reaction G, resulting in the observed difference between G and H. For the other reactions, the net change in entropy is none (as in answer D and C) or smaller (in B), so less of a difference is observed between G and H-TS.
 
I'm going to make an attempt at this question but not sure if this is right:

So Gibbs Free Energy equation says that delta G = delta H - TdeltaS?
In the first equation, 3 moles of material are combining to form product, which means that the change in entropy is negative (or not favored). By rearranging G+TS = H, the entropy term influences the net the energy of reaction G, resulting in the observed difference between G and H. For the other reactions, the net change in entropy is none (as in answer D and C) or smaller (in B), so less of a difference is observed between G and H-TS.

Thanks, I get why it's not C or D now- 0 entropy change. But in A there's 3 moles --> 2, and in B there's 2 moles --> 1, so how do we know the answer is A?
 
Thanks, I get why it's not C or D now- 0 entropy change. But in A there's 3 moles --> 2, and in B there's 2 moles --> 1, so how do we know the answer is A?

Enthalpy is an extensive property and in 'A' you go from 1 to 2 moles of GAS and in 'B' there is a 1 to 1 mole of GAS. Since the coefficients increase the enthalpy change increases.
 
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