Question

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

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

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.

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.