Spontaneous and non spontaneous

Question

The reaction between ammonia and dinitrogen pentoxide to give nitrogen gas and water, NH3(g) N2O3(g) <=> N2(g) H2O(g) has a DH value of -600.0 kJ and a DS value of -6 kJ/mol K. At what temperature does the reaction become non-spontaneous?

a. 100 oC

b. -173 oC

c. 100,000 K

d. -173 K

e. 0.1 oC

can you explain why the answer is B?

Answer 1

no- this question doesn&#039;t even make sense 2 me. But i&#039;ll *star* it anyway :)  

Answer 2

First off, you need DG, the total free energy change, to be negative for the reaction to be spontaneous.  For constant temperature, DG is defined as:

DG = DH - T*DS

DH = -600 kJ/mol

DS = -6 kJ/mol-K

Since DS is negative, there is a decrease in entropy for the reaction, meaning that -T*DS is positive and that as temperature increases, -T*DS gets bigger.  This is bad since you want DG to be less than zero for a spontaneous reaction.

Ok, so the temperature for b. is -173 C which is 100 K.  

-T*DS (@ T=100K) = 6 * 100 = 600

DG = -600 + 600 = 0

For T&gt; 100 K (=-173 C), DG &gt; 0 and the reaction isn&#039;t spontaneous.

p.s.  I think you have the units on DS wrong, it should be just kJ/K, not kJ/(mol-K)