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Step-by-Step Solution
Step 1: Identify the Reaction and Given Equilibrium Constant
We are given the reaction:
N2(g) + 3H2(g) ⇌ 2NH3(g)
The equilibrium constant for this reaction is given by $K_p$.
When pure ammonia (NH3) is allowed to dissociate, we consider the reverse process:
2NH3(g) ⇌ N2(g) + 3H2(g)
The equilibrium constant for this reverse reaction is $1/K_p$.
Step 2: Define Partial Pressures at Equilibrium
Let the initial pressure of ammonia (NH3) be $P^o$. Since only ammonia is present initially, its partial pressure at time t=0 is $P^o$, and that of N2 and H2 is 0. Suppose at equilibrium the partial pressure of ammonia is $P_{NH_3}$, that of N2 is $p$, and that of H2 is $3p$.
2NH3(g)
⇌
N2(g)
+
3H2(g)
At t = 0
Po
0
0
At t = teq
PNH3
p
3p
Step 3: Express the Total Pressure
The total pressure at equilibrium is:
$P_{\text{Total}} = P_{NH_3} + p + 3p = P_{NH_3} + 4p$
We are told $P_{NH_3}$ is much smaller than $P_{\text{Total}}$, so we neglect $P_{NH_3}$ in comparison to $4p$. Hence:
$P_{\text{Total}} \approx 4p$
Therefore:
$p = \dfrac{P_{\text{Total}}}{4}$
Step 4: Write the Equilibrium Constant Expression
For the reverse reaction 2NH3 ⇌ N2(g) + 3H2(g):
$K_{\text{eq}} = \dfrac{p_{N_2} \times (p_{H_2})^3}{(p_{NH_3})^2} = \dfrac{1}{K_p}$
Substituting $p_{N_2} = p$ and $p_{H_2} = 3p$ gives:
$\dfrac{1}{K_p} = \dfrac{p \times (3p)^3}{(p_{NH_3})^2} = \dfrac{27p^4}{(p_{NH_3})^2}$
Step 5: Solve for the Partial Pressure of NH3
Rearrange to find $(p_{NH_3})^2$:
$(p_{NH_3})^2 = 27\,p^4 \, K_p$
Taking the square root on both sides gives:
$p_{NH_3} = \sqrt{27} \,\sqrt{K_p}\, p^2 = 3^{3/2}\,\sqrt{K_p}\,p^2$
Step 6: Substitute $p = \dfrac{P_{\text{Total}}}{4}$
Since $p^2 = \left(\dfrac{P_{\text{Total}}}{4}\right)^2 = \dfrac{P_{\text{Total}}^2}{16}$, we get:
$p_{NH_3} = 3^{3/2} \,\sqrt{K_p}\,\dfrac{P_{\text{Total}}^2}{16}
= \dfrac{3^{3/2}\,\sqrt{K_p}\,P_{\text{Total}}^2}{16}$
This matches the correct expression given in the problem statement.
Final Answer
$p_{NH_3} = \dfrac{3^{3/2}\,\sqrt{K_p}\,P^2}{16}$
