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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

Expert verified

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

Step by step solution

01

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

02

Calculate water vapor pressure at equilibrium:

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

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A large piece of jewelry has a mass of 132.6g. A graduated cylinder initially contains 48.6 mL water. When the jewellery is submerged in the graduated cylinder, the total volume increases to 61.2 mL.

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