In a reaction \(2A \longrightarrow B\), the concentration of \(A\) decreases from \(0.64\, \ M\) to \(0.22\, \ M\) in \(5\) minutes. What is the rate of reaction (in \(Ms^{-1}\)) during this interval?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(7 \times 10^{-4}\, \ Ms^{-1}\)
The rate of a reaction must have one value whichever species it is measured by. When a reactant and a product appear with different stoichiometric coefficients, each rate of change is divided by its coefficient. For 2A → B, two moles of A are used up for every mole of B formed, so
\( \displaystyle r_{av} = -\frac{1}{2}\,\frac{\Delta[A]}{\Delta t} = \frac{\Delta[B]}{\Delta t} \)
The data give \( \Delta[A] = 0.22 - 0.64 = -0.42\,\mathrm{M} \) and \( \Delta t = 5\,\mathrm{min} = 300\,\mathrm{s} \).
\( \displaystyle r_{av} = -\frac{1}{2} \times \frac{(-0.42\,\mathrm{M})}{300\,\mathrm{s}} \)
\( \displaystyle r_{av} = \frac{0.42}{600}\,\mathrm{M\,s^{-1}} = 7 \times 10^{-4}\,\mathrm{M\,s^{-1}} \)
Note: NTA picked this question from an NCERT (2026-27) intext question and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Intext Questions – “3.2 In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L–1 to 0.4 mol L–1 in 10 minutes. Calculate the rate during this interval?”