
Given $2NO+{{O}_{2}}\to 2N{{O}_{2}}\text{ rate = K }\!\![\!\!\text{ NO}{{\text{ }\!\!]\!\!\text{ }}^{2}}[{{O}_{2}}]$ by how many times does the rate of reaction change when the volume of the reaction vessel is reduced to 1/3 of its original volume? Will there be any change in the order of the reaction?
Answer
577.8k+ views
Hint: Rate is defined as the speed at which a chemical reaction occurs. Rate is generally expressed in the terms of concentration of reactant which is consumed during the reaction in a unit of time or the concentration of product which is produced during the reaction in a unit of time.
Complete step by step solution:
Given in the question:
$2NO+{{O}_{2}}\to 2N{{O}_{2}}\text{ rate = K }\!\![\!\!\text{ NO}{{\text{ }\!\!]\!\!\text{ }}^{2}}[{{O}_{2}}]$
Now it is given that the reaction vessel is reduced to 1/3 of the original volume so that means that the concentration of the reaction also be reduced to 1/3 of the initial concentration.
So the new rate can be mentioned as :
$\begin{align}
& {{\text{R}}_{new}}\text{= K }\!\![\!\!\text{ NO/3}{{\text{ }\!\!]\!\!\text{ }}^{2}}[{{O}_{2}}/3]=K\dfrac{{{[NO]}^{2}}}{9}.\dfrac{[{{O}_{2}}]}{3} \\
& =\dfrac{1}{27}K{{[NO]}^{2}}[{{O}_{2}}] \\
\end{align}$
If we compare the initial rate and the new rate after reducing the concentration to 1/3 of the original value, we can say that the rate of the reaction is $\dfrac{1}{27}$ times the initial rate of the reaction.
There will be no change in the order of the reaction because the order of the reaction does not depend on the concentration of the reactants.
Hence the answer is
The rate of reaction change when the volume of the reaction vessel is reduced to 1/3 of its original volume= is $\dfrac{1}{27}$ times the initial rate of the reaction and there is no change in the order of the reaction
Note: If the reaction is a third order reaction, the unit for third order reaction is ${{M}^{-2}}h{{r}^{-1}}or\text{ mo}{{\text{l}}^{-2}}{{L}^{2}}h{{r}^{-1}}$. The negative and positive sign in the expression of the rate or reaction only means the change in concentration. A negative charge indicates that the concentration of the reactant is decreasing, similarly a positive charge means that the concentration of product is increasing
Complete step by step solution:
Given in the question:
$2NO+{{O}_{2}}\to 2N{{O}_{2}}\text{ rate = K }\!\![\!\!\text{ NO}{{\text{ }\!\!]\!\!\text{ }}^{2}}[{{O}_{2}}]$
Now it is given that the reaction vessel is reduced to 1/3 of the original volume so that means that the concentration of the reaction also be reduced to 1/3 of the initial concentration.
So the new rate can be mentioned as :
$\begin{align}
& {{\text{R}}_{new}}\text{= K }\!\![\!\!\text{ NO/3}{{\text{ }\!\!]\!\!\text{ }}^{2}}[{{O}_{2}}/3]=K\dfrac{{{[NO]}^{2}}}{9}.\dfrac{[{{O}_{2}}]}{3} \\
& =\dfrac{1}{27}K{{[NO]}^{2}}[{{O}_{2}}] \\
\end{align}$
If we compare the initial rate and the new rate after reducing the concentration to 1/3 of the original value, we can say that the rate of the reaction is $\dfrac{1}{27}$ times the initial rate of the reaction.
There will be no change in the order of the reaction because the order of the reaction does not depend on the concentration of the reactants.
Hence the answer is
The rate of reaction change when the volume of the reaction vessel is reduced to 1/3 of its original volume= is $\dfrac{1}{27}$ times the initial rate of the reaction and there is no change in the order of the reaction
Note: If the reaction is a third order reaction, the unit for third order reaction is ${{M}^{-2}}h{{r}^{-1}}or\text{ mo}{{\text{l}}^{-2}}{{L}^{2}}h{{r}^{-1}}$. The negative and positive sign in the expression of the rate or reaction only means the change in concentration. A negative charge indicates that the concentration of the reactant is decreasing, similarly a positive charge means that the concentration of product is increasing
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