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Question: Given \(2NO + {O_2} \to 2N{O_2}\), \({\text{Rate}} = K{\left[ {NO} \right]^2}\left[ {{O_2}} \righ...

Given 2NO+O2→2NO22NO + {O_2} \to 2N{O_2},
Rate=K[NO]2[O2]{\text{Rate}} = K{\left[ {NO} \right]^2}\left[ {{O_2}} \right] by how many times does the rate of reaction change when the volume of the reaction vessel is reduced to 1/3rd1/{3^{rd}}of its original volume? Will there be any change in the order of the reaction?

Explanation

Solution

To answer this question, you must recall the formula for rate of a reaction. The rate of a reaction varies with the concentration of reactants depending upon the order of reaction.
Formula used:
rate = k[A]a[B]b{\text{rate = k}}{\left[ {\text{A}} \right]^{\text{a}}}{\left[ {\text{B}} \right]^{\text{b}}}
Where a and b are the stoichiometric coefficients and [A] and [B]\left[ {\text{A}} \right]{\text{ and }}\left[ {\text{B}} \right] are the molarity of the reactants.

Complete step by step answer:
For the given reaction, 2NO+O2→2NO22NO + {O_2} \to 2N{O_2}
The rate law can be given as, Rate = k[NO]2[O2]{\text{Rate = k}}{\left[ {{\text{NO}}} \right]^{\text{2}}}\left[ {{{\text{O}}_{\text{2}}}} \right].
We can write concentration in terms of number of moles and volume.
Thus, [NO] = nV\left[ {{\text{NO}}} \right]{\text{ = }}\dfrac{{\text{n}}}{{\text{V}}}
So, we can write the rate expression as, R = k[nV]2[nV]{\text{R = k}}{\left[ {\dfrac{{\text{n}}}{{\text{V}}}} \right]^{\text{2}}}\left[ {\dfrac{{\text{n}}}{{\text{V}}}} \right].
We are given in the question, that, volume of the vessel is reduced to 1/3rd1/{3^{rd}}of its original value,
⇒V = V3\Rightarrow {\text{V = }}\dfrac{{\text{V}}}{{\text{3}}}
Thus, the concentration becomes,
[NO] = nV/3=3nV\left[ {{\text{NO}}} \right]{\text{ = }}\dfrac{{\text{n}}}{{{\text{V/3}}}} = \dfrac{{3{\text{n}}}}{{\text{V}}}.
The new rate expression can be written as,
R’ = k[3nV]2[3nV]{\text{R' = k}}{\left[ {\dfrac{{{\text{3n}}}}{{\text{V}}}} \right]^{\text{2}}}\left[ {\dfrac{{{\text{3n}}}}{{\text{V}}}} \right]
⇒R’ = k(27)[nV]2[nV]\Rightarrow {\text{R' = k}}\left( {{\text{27}}} \right){\left[ {\dfrac{{\text{n}}}{{\text{V}}}} \right]^{\text{2}}}\left[ {\dfrac{{\text{n}}}{{\text{V}}}} \right]
Comparing the new rate with the original rate expression we get, R’=27R{\text{R'}} = 27{\text{R}}
Thus, the rate of reaction increases by 27 times.
We can see in the new rate law expression that the powers of the concentration of reactants remain the same. Hence the order of the reaction is 2+1=32 + 1 = 3, which is the same as earlier.
Thus, we can conclude that the order of a reaction does not change with change in the volume of the vessel.
Therefore, we can say that there will be no change in the order of the reaction.

Note:
The rate law is an experimentally determined quantity and can be used to predict the relationship between the concentrations of reactants and the rate of the reaction. For elementary reactions, the rate equation can be simply derived from first principles using the collision theory. The rate equation of a reaction with a multi-step mechanism cannot be calculated from the stoichiometric coefficients of the overall reaction.