Question

Increase in the pressure for the following equilibrium: ${H_2}O\left( l \right) \rightleftharpoons {H_2}O\left( g \right)$, result in:
A.Formation of more ${H_2}O\left( l \right)$
B.Formation of more ${H_2}O\left( g \right)$
C.Increase in boiling point of ${H_2}O\left( l \right)$
D.Decrease in boiling point of ${H_2}O\left( l \right)$

Answer 1

Hint: We can predict the formation of reactants and products in an equilibrium reaction using Le Chatelier’s principle. Pressure and boiling point are proportional to each other.

Complete step by step answer:
Le-Chatelier’s Principle:
At equilibrium, if a chemical system is disturbed, the system would react in the direction which counteracts the disturbance and the stress.
Example:
$C{O_{\left( g \right)}} + {H_2}{O_{\left( l \right)}} \rightleftarrows C{O_2}_{\left( g \right)} + {H_{2\left( g \right)}}$
If the concentration of the reactant is increased the equilibrium of the reaction shifted to the right side.
If the concentration of the reactant is decreased the equilibrium of the reaction shifted to the left side
Based on Le-Chatelier’s Principle, when we increase the pressure, the shift in the equilibrium would take place, so that pressure would decrease again. The striking of the gas molecules in the sides of the container causes pressure. The pressure will be higher, if the number of gaseous molecules is higher.
When it comes to pressure changes, increasing gaseous pressure would shift the equilibrium towards the side that contains a lesser number of the gases molecules. If we decrease the pressure, the equilibrium would shift in the direction that contains more number of gaseous molecules.
The given equilibrium reaction is,
${H_2}O\left( l \right) \rightleftharpoons {H_2}O\left( g \right)$
In this reaction, the effect of pressure would be the shift in the equilibrium towards the backward direction. This is based on Le Chatelier’s Principle that states the equilibrium would move in the direction of fewer molecules of gaseous.
In the given reaction, the number of gas molecules in the reactant side is zero, and the number of gas molecules in the product side is one. So, when we increase the pressure, the equilibrium would position in the direction of fewer molecules of gas. The equilibrium would favor the reactant side i.e. formation of more ${H_2}O\left( l \right)$ could take place.

We know that as the pressure increases, the boiling point of liquid also increases. This is because boiling point is the point where the vapour pressure is equal (or) is above the atmospheric pressure. Therefore, when we increase the pressure, the boiling point of the ${H_2}O\left( l \right)$ also increases.

So, the correct answer is Option A,C.

Note:
We have to know that few factors that affect the shift in equilibrium are,
Changes in concentration
Effect of change in temperature on equilibrium
Effect of catalysts on equilibrium

For example: The direction of the equilibrium shift when the concentration of hydrogen chloride is increased has to be given.
The given reaction is,
${H_2}\left( g \right) + C{l_2}\left( g \right) \rightleftarrows 2HCl\left( g \right)$
According to Le-Chatelier’s principle, increasing the concentration of hydrogen chloride shifts the equilibrium to the left to form more reactants.