Complete Question
The complete question is shown on the first uploaded image
Answer:
The concentration equilibrium constant is 
Explanation:
The chemical equation for this decomposition of ammonia is
↔ 
The initial concentration of ammonia is mathematically represented a
![[NH_3] = \frac{n_1}{V_1} = \frac{29}{75}](https://tex.z-dn.net/?f=%5BNH_3%5D%20%3D%20%20%5Cfrac%7Bn_1%7D%7BV_1%7D%20%20%3D%20%5Cfrac%7B29%7D%7B75%7D)
![[NH_3] = 0.387 \ M](https://tex.z-dn.net/?f=%5BNH_3%5D%20%3D%200.387%20%20%5C%20%20M)
The initial concentration of nitrogen gas is mathematically represented a
![[N_2] = \frac{n_2}{V_2}](https://tex.z-dn.net/?f=%5BN_2%5D%20%3D%20%20%5Cfrac%7Bn_2%7D%7BV_2%7D)
![[N_2] = 0.173 \ M](https://tex.z-dn.net/?f=%5BN_2%5D%20%3D%20%200.173%20%20%5C%20%20M)
So looking at the equation
Initially (Before reaction)


During reaction(this is gotten from the reaction equation )
(this implies that it losses two moles of concentration )
(this implies that it gains 1 moles)
(this implies that it gains 3 moles)
Note : x denotes concentration
At equilibrium


Now since
![[NH_3] = 0.387 \ M](https://tex.z-dn.net/?f=%5BNH_3%5D%20%3D%200.387%20%20%5C%20%20M)
Now the equilibrium constant is
![K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=K_c%20%20%3D%20%20%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
substituting values

