Question

This is the equation for the dissociation of ammonia gas at 293 K. Delta. H = 145 kJ and Delta. S = 195 J/k. 2NH3(g) Right arrow. N2(g) 3H2(g) Which correctly states the Delta. G for this dissociation and whether the process is spontaneous or nonspontaneous? Use Delta. G = Delta. H – TDelta. S. –87. 87 kJ, spontaneous –50 kJ, spontaneous 87. 9 kJ, nonspontaneous 202. 14 kJ, nonspontaneous.

Answer 1

The delta G for the dissociation is 87,865 joule/kg.

The process is non-spontaneous because the value of Gibbs free energy is positive.

What is dissociation?

The breaking up of a compound into a simpler form that can be recombined under other conditions.

Given,

$\bold{2NH_3 (g) = N_2(g) + 3H_2(g)}$

H = 145 kJ

Delta S = 195 J/k

Determining the G using the following process:

G = H -TS

$\bold{G = (145\times10^3 J) -(293K \times 195 J/K})$

G = 87,865 Joules/K

Thus, the value of Delta G is  87,865 and the process is non-spontaneous.

Learn more about dissociation here:

https://brainly.com/question/4431132

Answer 2

Answer:

C. 87.9 kJ, nonspontaneous

Explanation:

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