Question

Sodium sulfate 10−hydrate, \(N{a_2}S{O_4}.10{H_2}O\), dehydrates according to the equation

\(C(s) + {H_2}O(g) \rightleftharpoons CO(g) + {H_2}(g)\)

\({K_P} = 4.08 \times 1{0^{ - 25}}at2{5^o}C\)

What is the pressure of water vapor at equilibrium with a mixture of \(N{a_2}S{O_4}.10{H_2}O\)and \(NaS{O_4}\)?

Short answer

The vapor pressure of water at equilibrium is \(3.64x1{0^{ - 3}}atm\).

Step-by-step solution

Given information:

1. Value of equilibrium constant at \(2{5^0}C\)is \({K_P} = 4.08 \times 1{0^{ - 25}}\)

The value of equilibrium water vapor pressure needs to be calculated

Calculate water vapor pressure at equilibrium:

\(\begin{array}{*{20}{c}}{{K_p}}&{ = {{\left( {{P_{{H_2}O}}} \right)}^{10}}}\\{{P_{{H_2}O}}}&{ = \sqrt[{10}]{{{K_p}}}}\\{}&{ = \sqrt[{10}]{{4.08 \times {{10}^{ - 25}}}}}\\{}&{}\\{}&{ = 3.64 \times {{10}^{ - 3}}{\rm{atm}}}\end{array}\)