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At a temperature of 60 ̊C, the vapor pressure of water is 0.196atm. What is the value of the equilibrium constant Kp for the transformation at 60 ̊C? H2O (l)⇌ H2O(g)

Short Answer

Expert verified

The value of the equilibrium constant Kp=0.196

Step by step solution

01

Define the vapor pressure of water

The vapor pressure of water is the pressure exerted by molecules of water vapor in gaseous form. The saturation vapor pressure is the pressure at which water vapor is in thermodynamic equilibrium with its condensed state.

02

Calculating  Kp

The reaction H2O (l) ⇌ H2O(g)

The vapor pressure of water is \({P_{{H_2}O(g)}}=0.196{\rm{atm}}\)

Solids and liquids are not included in the equilibrium constant expression

The value of Kp=0.196

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Most popular questions from this chapter

If you observe the following reaction at equilibrium, is it possible to tell whether the reaction stated with pure \(N{O_2}\) or with pure \({N_2}{O_4}\)? \(2N{O_2}(g) \rightleftharpoons {N_2}{O_4}(g)\)

Write the expression of the reaction quotient for the ionization of HOCN in water.

Convert the values of Kc to values of Kp or the values of Kp to values of Kc .

\((a)\,{N_2}\left( g \right) + 3{H_2}\left( g \right)\rightleftharpoons 2N{H_3}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_C} = 0.50\,\,at\,\,400^\circ C\)

\((b){{\rm{H}}_2}(g) + {{\rm{I}}_2}(g)\rightleftharpoons 2HI(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_c} = 50.2\,at\,{448^\circ }{\rm{C}}\)

\((c)N{a_2}{\rm{S}}{{\rm{O}}_4} \cdot 10{{\rm{H}}_2}O(s)\rightleftharpoons N{a_2}{\rm{S}}{{\rm{O}}_4}(s) + 10{{\rm{H}}_2}O(g){K_P} = 4.08 \times {10^{ - 25}}at\,{25^\circ }{\rm{C}}\)

\((d){{\rm{H}}_2}{\rm{O}}(l)\rightleftharpoons {{\rm{H}}_2}{\rm{O}}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_P} = 0.122\,at\,{50^\circ }{\rm{C}}\)

Calculate the pressures of all species at equilibrium in a mixture of NOCl, NO, and Cl2produced when a sample of NOCl with a pressure of 10.0 atm comes to equilibrium according to this reaction:

\(2NOCl(g) \rightleftharpoons 2NO(g) + C{l_2}(g)\quad {K_P} = 4.0 \times 1{0^{ - 4}}\)

How can the pressure of water vapor are increased in the following equilibrium?

\({H_2}O(l) \rightleftharpoons {H_2}O(g)\) \(\Delta H = 41kJ\)

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