Equilibrium Question 733
Question: For the reaction $ NH_4HS(g)\rightarrow NH_3(g)+H_2S(g) $ in a closed flask, the equilibrium pressure is P atm. The standard free energy of the reaction would be:
Options:
A) $ -RT $ ln p
B) $ -RT $ (ln p - ln 2)
C) $ -2RT $ ln p
D) $ -2RT $ (ln p - ln 2)
Show Answer
Answer:
Correct Answer: D
Solution:
$ {P_{NH_3}}={P_{H_2S}}=\frac{P}{2}atm $
$ K_{p}={P_{NH_3}}{P_{H_2S}}={{( \frac{P}{2} )}^{2}}=\frac{P^{2}}{4} $
$ \Delta G=-RT,ln,K_{p}=-RT,ln,{{( \frac{p}{2} )}^{2}} $
$ =-2RT[ ln,p-ln,2 ] $
BETA
