1. Write the equilibrium reaction and expression for $K_p$

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

So, $K_p=\frac{(P_{NH_3})^2}{P_{N_2}(P_{H_2})^3}$

2. Start with pure ammonia and let it dissociate

Since initially only ammonia is present, the reverse reaction occurs: $2NH_3 \rightleftharpoons N_2+3H_2$

Let at equilibrium the partial pressures formed be:

• $P_{N_2}=x$
• $P_{H_2}=3x$

Then ammonia consumed corresponds to $2x$, so if total pressure at equilibrium is $P$, and ammonia partial pressure is small compared to total pressure, we use: $P_{NH_3}\ll P$

Hence most of the total pressure comes from $N_2$ and $H_2$.

Now, $P=P_{N_2}+P_{H_2}+P_{NH_3}\approx x+3x=4x$

So, $x\approx \frac{P}{4}$

Thus, $P_{N_2}=\frac{P}{4},\qquad P_{H_2}=\frac{3P}{4}$

3. Substitute into the equilibrium constant expression

Using $K_p=\frac{(P_{NH_3})^2}{P_{N_2}(P_{H_2})^3}$

we get $K_p=\frac{(P_{NH_3})^2}{\left(\frac{P}{4}\right)\left(\frac{3P}{4}\right)^3}$

Now simplify the denominator: $\left(\frac{P}{4}\right)\left(\frac{27P^3}{64}\right)=\frac{27P^4}{256}$

Therefore, $K_p=\frac{(P_{NH_3})^2}{27P^4/256}$

So, $(P_{NH_3})^2=K_p\cdot \frac{27P^4}{256}$

Taking square root: $P_{NH_3}=\frac{\sqrt{27}\,K_p^{1/2}P^2}{16}$

Since $\sqrt{27}=3^{3/2}$

we obtain: $P_{NH_3}=\frac{3^{3/2}K_p^{1/2}P^2}{16}$

4. Match with the options

This matches Option C: $\boxed{\frac{3^{3/2}K_p^{1/2}P^2}{16}}$

5. Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

So the derived answer agrees with the stored answer.