Question

The equation: $log\frac{k_c}{k_P}-log\frac{1}{RT}=0$, is applicable to which of the following equilibrium reactions?

• A. $N{H}_3\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)$
• B. $CaC{O}_3\left(s\right)\rightleftharpoons Ca{O}_3\left(s\right)+C{O}_2\left(g\right)$
• C. $2N{O}_2\left(g\right)\rightleftharpoons N_2{O}_4\left(g\right)$
• D. $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$

Answer 1

Answer: (A), (B)

$N{H}_3\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)$
$CaC{O}_3\left(s\right)\rightleftharpoons Ca{O}_3\left(s\right)+C{O}_2\left(g\right)$

The relationship between the two equilibrium constants are: $K_p=K_c(RT{)}^{\Delta n}$
The given equation is: $log\frac{k_c}{k_P}-log\frac{1}{\left(RT\right)}=0$ $⟹k_p=k_c\left(RT\right)$ $⟹\Delta n=1$.
In option A, $\Delta n=(\frac{1}{2}+\frac{3}{2})-1=1$.
In option B, $\Delta n=1-0=1$.
In option C, $\Delta n=1-2=-1$.
in option D, $\Delta n=2-2=0$.