Question

The reaction is of the first order. If the volume of reaction vessel is reduced to 1/3, the rate of the reaction would be:
A. \[\dfrac{1}{3}\] times
B. \[\dfrac{2}{3}\] times
C. \[3\] times
D. \[6\] times

Answer 1

Hint: The equilibrium constant of a reaction is the equilibrium concentration of the products by the equilibrium concentration of reactants and the exponentiation of concentration is determined by the number of moles of the reactants and products. Concentration is given by moles of solute divided by the volume in litres. When the volume is reduced by \[\dfrac{1}{3}\] then it affects the concentration.

Complete answer:
First we need to find the effect on concentration when the volume is reduced to \[\dfrac{1}{3}\] , the equation for concentration denoted by C can be written as
\[{\text{C = }}\dfrac{{{\text{mol}}}}{{\text{V}}}\]
Where mol is the number of moles and V is the volume in litres.
The volume is inversely proportional to the concentration. So if the volume is decreased then the concentration increases but if the volume is increased then the concentration decreases.
The volume is reduced to \[\dfrac{1}{3}\] ,then the equation becomes
\[\dfrac{{{\text{mol}}}}{{{\text{V/3}}}}\] , the 3 gets into numerator and we get
\[\dfrac{{3{\text{mol}}}}{{\text{V}}}\] thus, substituting the \[\dfrac{{{\text{mol}}}}{{\text{V}}}\] by \[{\text{C}}\] we get \[{\text{3C}}\]
The equilibrium concentration of the given reaction can be written as
\[{\text{K = }}\dfrac{{{{[{\text{N}}{{\text{O}}_2}]}^2}}}{{{{[{\text{NO]}}}^2}[{{\text{O}}_2}]}}\]
That is the concentration of the products by the concentration of reactants at equilibrium. The exponentiation is the number of moles present in the product or reactant.
Now replacing the concentration of the reactants and products by \[{\text{3C}}\]
We get the equilibrium constant as,
\[{\text{K = }}\dfrac{{{{{\text{[3C]}}}^2}}}{{{{[{\text{3C]}}}^2}[{\text{3C]}}}}\]
The \[{[3{\text{C]}}^2}\] can be cancelled from both the numerator and denominator.
Thus we are left with \[\dfrac{1}{3}\] ,that is the rate of the reaction would be \[\dfrac{1}{3}\] times when the volume of the reaction vessel is reduced to \[\dfrac{1}{3}\] .
And hence the answer is option A.

Note:
Remember the equation of equilibrium constant. On understanding the relationship between concentration and volume, the reduction is only applicable to volume. Since both are inversely proportional when one component increases the other decreases and vice versa. On calculating the equilibrium concentration, the number values can be taken out, in this case it will turn 9, as the concentration is squared. But the final answer remains the same.