Question

In the reaction $H_2\left(g\right)+{Cl}_2\left(g\right)\rightleftharpoons 2HCl\left(g\right)$:

• A. $K_p\ne K_c$
• B. $K_p=K_c$
• C. $K_p>K_c$
• D. $K_p<K_c$

Answer 1

Answer: (B)

$K_p=K_c$

$K_p=K_c$

$H_2\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons 2HCl\left(g\right)$

$Kc=\frac{[HCl{]}^2}{\left[H_2\right]\left[C{l}_2\right]}$

Assuming all the gases behave ideally, we can write $P_{i}V=n_{i}RT$ for each gas but now look at the units of concentration, which are mol/L. That is just means that $\frac{n_i}{V}=\frac{P_i}{RT}$.

Substitute that into Kc:

$K_c=\frac{{\frac{P_{\left(HCl\right)}}{RT}}^2}{\frac{P_{\left(H_2\right)}}{RT}\times \frac{P_{\left(C{l}_2\right)}}{RT}}$

$=\frac{P_{\left(HCl\right)}^2}{P_{\left(H_2\right)}\times P_{\left(C{l}_2\right)}}=K_p$