Question

The value of $K_c$ is 64 at 800 K for the reaction: $N_2 + 3H_2 \rightleftharpoons 2NH_3$ (all gases). Therefore the value of $K_c$ at 800 K for the reaction $NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2$ (all gases) is

• A. $\frac{1}{64}$
• B. 8
• C. $\frac{1}{4}$
• D. $\frac{1}{8}$

Answer 1

Answer: (D)

$\frac{1}{8}$

Answer 2

To find the value of $K_c$ for the reaction $NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2$, we need to relate it to the given reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$.

Let the given reaction be (1):

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

The equilibrium constant for reaction (1) is $K_{c1} = 64$. The expression for $K_{c1}$ is:

$K_{c1} = \frac{[NH_3]^2}{[N_2][H_2]^3} = 64$

Let the target reaction be (2):

$NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)$

We need to find $K_{c2}$. The expression for $K_{c2}$ is:

$K_{c2} = \frac{[N_2]^{1/2}[H_2]^{3/2}}{[NH_3]}$

Comparing reaction (2) with reaction (1), we can observe two changes:

1. Reaction (2) is the reverse of reaction (1).
2. The stoichiometric coefficients of reaction (2) are half of the stoichiometric coefficients of the reverse of reaction (1).

Step 1: Reverse reaction (1).

If a reaction is reversed, the new equilibrium constant is the reciprocal of the original equilibrium constant. Let's reverse reaction (1) to get reaction (1'):

$2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$

The equilibrium constant for reaction (1'), $K_{c1}'$, will be:

$K_{c1}' = \frac{1}{K_{c1}} = \frac{1}{64}$

Step 2: Divide the stoichiometric coefficients of reaction (1') by 2.

If the stoichiometric coefficients of a reaction are multiplied by a factor 'n', the new equilibrium constant is the original equilibrium constant raised to the power 'n'. In this case, we are multiplying by $n = \frac{1}{2}$ (which is equivalent to taking the square root).

Reaction (2) is obtained by multiplying reaction (1') by $\frac{1}{2}$:

$\frac{1}{2} (2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g))$

This gives:

$NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)$

The equilibrium constant for reaction (2), $K_{c2}$, will be:

$K_{c2} = (K_{c1}')^{1/2}$

$K_{c2} = \left(\frac{1}{64}\right)^{1/2}$

$K_{c2} = \sqrt{\frac{1}{64}}$

$K_{c2} = \frac{\sqrt{1}}{\sqrt{64}}$

$K_{c2} = \frac{1}{8}$

The value of $K_c$ at 800 K for the reaction $NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2$ is $\frac{1}{8}$.