Question

For the reaction of $1.0M{\text{ }}HCl$ with $1.0M{\text{ }}NaOH$ in a closed system at ${25.0^o}C,$ how is the rate affected when the pressure is increased by 50%?

Answer 1

Hint: When there is an increase in pressure the equilibrium will shift towards the side of the reaction with fewer moles of a gas. When there is decrease in pressure equilibrium will shift to the side of the reaction with more moves of gas and no change of pressure on equilibrium and solid constituents.

Complete step by step answer:
$NaO{H_{\left( l \right)}} + HC{L_{\left( e \right)}} \to NaC{I_{\left( s \right)}} + {H_2}{O_{\left( l \right)}}$
There are no gaseous reactant molecules so if there are no gaseous reactant molecules then pressure has no effect on the rate of reaction because liquids and solids are almost impossible to compress to increase the concentration.
So it doesn’t put any effect on the rate of collision determining the speed of the reaction.
So the correct option is (D) the rate does change as there are only liquid reactants.

Additional Information
Collision involving two particles
The same argument applies whether the reaction invoices colliding between two different articles are two of the same particles.
In order for any reaction to happen those parties most first collide this is true whether loath particles are in gas state or whether one is a gas and other is a solid. If the pressure is higher the chances of collision are greater.

Pressure only affects rate when the reactants are in the gaseous state.

Note:
Factors that influence the reaction rates of chemical reactions include the concentration of reactants, temperature, the physical state of reactants and their dispersion, and the presence of a catalyst.