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

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

Short Answer

Expert verified

The pressure of water vapor increased in the above-mentioned equilibrium by

(i) Increase in temperature

(ii) Increase in the amount of \({{\rm{H}}_2}{\rm{O}}({\rm{l}})\).

Step by step solution

01

Understanding equilibrium

A physical change

\({\rm{\Delta H}} = 41{\rm{k}}.{\rm{J}}\)

Equilibrium can be established for a physical change as well as for a chemical reaction.

02

Increasing pressure of water vapor

Let us see how the pressure of water vapor can be increased.

(i) Since \({\rm{\Delta H}} > 0\)the reaction is endothermic, hence

\({H_2}O(l) \rightleftharpoons {H_2}O(g)\)

By increasing the temperature (heat), the equilibrium of a physical change will move to the right and the amount of \({{\rm{H}}_2}{\rm{O}}({\rm{g}})\) will increase; hence the pressure of water vapor will increase.

(ii) By increasing the amount of\({{\rm{H}}_2}{\rm{O}}({\rm{l}})\), the equilibrium of a physical change will move to the right, and the amount of \({{\rm{H}}_2}{\rm{O}}({\rm{g}})\) will increase; hence the pressure of water vapor will increase.

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Most popular questions from this chapter

Calcium chloride 6−hydrate, \(CaC{l_2}.6{H_2}O\), dehydrates according to the equation

\(CaC{l_2} \times 6{H_2}O(s) \rightleftharpoons CaC{l_2}(s) + 6{H_2}O(g)\)

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What is the pressure of water vapor at equilibrium with a mixture of \(CaC{l_2}.6{H_2}O\)and \(CaC{l_2}\)?

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(d) If the change in the pressure of NO2 is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O2 change?

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Question:What is the value of the equilibrium constant at \(50{0^o}C\) for the formation of \(N{H_3}\)according to the following equation? N2(g) + 3H2(g) ⇌ 2NH3(g)

An equilibrium mixture of \(N{H_3}(g)\) \({H_2}(g)\) and \({N_2}(g)\) at \(50{0^o}C\) was found to contain\(1.35M{H_2},1.15M{N_2}\)and \(4.12\)\( \times 1{0^{ - 1}}MN{H_3}\)

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