Given $k_p$ for the reaction $\frac{1}{2} C(g) ightleftharpoons \frac{1}{2} A(g)+\frac{1}{2} B(g)$ at a…
Given $k_p$ for the reaction
$\frac{1}{2} C(g) ightleftharpoons \frac{1}{2} A(g)+\frac{1}{2} B(g)$ at a fixed temperature is $0.25 \mathrm{~atm}^{-2}$. Find the $K_p$ for the reaction
$A(g)+B(g) ightleftharpoons C(g)$ at the same temperature.
16
25
9
36
Solution
For the reaction,
$$
\frac{1}{2} C(g) ightleftharpoons \frac{1}{2} A(g)+\frac{1}{2} B(g)
$$
Given, $K_p=0.25 \mathrm{~atm}^{-2}$
We know that, $K_C=\frac{[\text { Product }]^b}{[\text { Reactant }]^a}$, where $a$ and $b$ are stiochiometric coefficients.
Also, $K_p=K_C(R T)^{\Delta n_g}$
where, $\Delta n_g=b-a$
For above reaction, $K_{C_1}=\frac{[A]^{0.5}[B]^{0.5}}{[C]^{0.5}}$
$$
\Delta n_g=1-0.5=0.5
$$
So,
$$
K_{p_1}=K_{C_1}(R T)^{0.5}...(i)
$$
Similarly for reaction,
$A(g)+B(g) ightleftharpoons C(g)$, at same temperature
$$
\begin{aligned}
K_{C_2} & =\frac{[C]}{[A][B]} \\
\Delta n_g & =1-2=-1 \\
K_{p_2} & =K_{C_2}(R T)^{-1}...(ii)
\end{aligned}
$$
Now, divide $K_{p_1}$ and $K_{p_2}$, we get,
$$
\frac{K_{p_1}}{K_{p_2}}=\frac{K_{C_1}[R T]^{0.5}}{K_{C_2}[R T]^{-1}} \Rightarrow K_{p_2}=16 \mathrm{~atm}
$$
So, $K_p$ for the reaction
$$
A(g)+B(g) ightleftharpoons C(g) \text { is } 16 \text { atm. }
$$