Question

The vapour pressure of water at ${25}^{o}C$ is 0.0313 atm. The value of $K_p$ and $K_c$ at ${25}^{o}C$ for the following equilibrium is $H_{2}O\left(l\right)\leftrightharpoons H_{2}O\left(g\right)$.

• A. $1.28\times {10}^{-3}$ and 0.0313 respectively
• B. $0.0313$ and $1.28\times {10}^{-3}$ respectively
• C. $1.28\times {10}^{-3}$ and $1.28\times {10}^{-3}$ respectively
• D. $0.313$ and $0.0313$ respectively

Answer 1

The correct option is B $0.0313$ and $1.28\times {10}^{-3}$ respectively

$K_p=P_{H_{2}O}=0.0313$.

But $K_p=K_c(RT{)}^{\Delta n}=K_c(RT)$ because $\Delta n=1$.

Substitute values,

$0.0313=K_c(0.0821\times 298)$
$K_c=1.28\times {10}^{-3}$.