2001 AIPMT Chemical KineticsRate law and rate constant Easy Question 2 of 2 in Rate law and rate constant

For the reaction $2N_2O_5 \rightarrow 4NO_2 + O_2$, rate and rate constant are $1.02 \times 10^{-4}$ and $3.4 \times 10^{-5}$ s$^{-1}$ respectively. Then conc. of $N_2O_5$ at that time will be : -

A 1.732
B 3
C $1.02 \times 10^{-4}$
D $3.4 \times 10^5$

The answer is B.

Solution

The unit of the rate constant (s$^{-1}$) shows that this is a first order reaction.
$\therefore$ rate $= k[N_2O_5]$
$[N_2O_5] = \dfrac{\text{rate}}{k} = \dfrac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}}$
$[N_2O_5] = 3$ mol L$^{-1}$

Concept Concentration from rate of first order reaction

Rate law and rate constant in past papers

2 questions from this chapter have appeared across 2 exam years.