Question

For the reaction, $2{{N}_{2}}{{O}_{5}}->4N{{O}_{2}}+{{O}_{2}}$, the rate equation can be expressed in two ways i.e \[-\dfrac{d[{{N}_{2}}{{O}_{5}}]}{dt}=k[{{N}_{2}}{{O}_{5}}]\] and \[+\dfrac{d[N{{O}_{2}}]}{dt}=k\left[ {{N}_{2}}{{O}_{5}} \right]\]. Here, k and k’ are related as:
A. $k=4k'$
B. $k=2k'$
C. $2k=k'$
D. $k=k'$

Answer 1

Hint: In the question, we have been given one reaction and two reactions involving the rate constant. Try to manipulate the three of them to find a relation between k and k’.

Complete step-by-step solution:
First, let us understand what rate constant means in chemical kinetics. The rate constant, or the specific rate constant, is the proportionality constant in the equation that is used to express the relationship between the rate and the concentrations of the reacting substances, of a chemical reaction. The rate constant is denoted by ‘k’. You need to remember that while equating the reactants and products with respect to time, the coefficient of the reactants or products gets down as the denominator. As the reactants reduce in concentration over time, they have a negative sign in front of them. Whereas product formation increases over time, so they are shown with a positive sign. Let us write the 3 equations and try to manipulate them:
We have,
$-\dfrac{d[{{N}_{2}}{{O}_{5}}]}{2dt}=+\dfrac{d[N{{O}_{2}}]}{4dt}$…….(1)
\[-\dfrac{d[{{N}_{2}}{{O}_{5}}]}{dt}=k[{{N}_{2}}{{O}_{5}}]\]……….(2)
And,
$+\dfrac{d[N{{O}_{2}}]}{dt}=k'[{{N}_{2}}{{O}_{5}}]$…………(3)
Equating the equations, we have,
$-\dfrac{d[{{N}_{2}}{{O}_{5}}]}{dt}=\dfrac{k[{{N}_{2}}{{O}_{5}}]}{2}=+\dfrac{d[N{{O}_{2}}]}{4dt}=\dfrac{k'[{{N}_{2}}{{O}_{4}}]}{4}$
So, we obtain
$\dfrac{k}{2}=\dfrac{k'}{4}$
So,
$2k=k'$
This is the resultant relation for k and k’ which happens to be our answer.

So, the correct option is C.

Note: If you are unable to proceed with the question, the first thing is to write down the three equations and observe them carefully. Most numericals of chemical kinetics deal with equations which can be manipulated i.e multiplication and division to get an answer. Also, check for the sign or else the answer might get wrong.