Question

Equilibrium constant for the following equilibrium is given at $0^{o}C$
$N{a}_{2}HP{O}_4.12H_{2}O\rightleftharpoons N{a}_{2}HP{O}_4.7H_{2}O\left(s\right)+5H_{2}O\left(g\right)$ $K_p=31.25\times {10}^{-23}$
at equilibrium what will be partial pressure of water vapour:

• A. $\frac{1}{5}\times {10}^{-3}atm$
• B. $0.5\times {10}^{-3}atm$
• C. $5\times {10}^{-2}atm$
• D. $5\times {10}^{-3}atm$

Answer 1

Answer: (A)

$\frac{1}{5}\times {10}^{-3}atm$

${N{a}_{2}HP{O}_4.12H_{2}O}_{\left(g\right)}\rightleftharpoons N{a}_{2}HP{O}_4.7H_{2}O\left(s\right)+5H_{2}O\left(g\right)$

$K_P={\left(P_{H_{2}O}\right)}^5$

Given, $K_p=31.25\times {10}^{-23}={\left(P_{H_{2}O}\right)}^5$

Partial pressure of water vapor, ${\left(P_{H_{2}O}\right)}^5=5\times {10}^{-5}$