During the decomposition of H₂O₂ to give oxygen, 48 g O₂ is formed per minute at a certain point of time. The rate of formation of water at this point is:

3.0 mol min^–1

Here, the given reaction is

\(2H_2O_2 −−−−→ 2H_2O  +  O_2\)

Rate \(= − \frac{1}{2}\frac{d[H_2O_2]}{dt} = \frac{1}{2}\frac{d[H_2O]}{dt} = \frac{d[O_2]}{dt}\)

Rate of formation of oxygen \(= 48 \text{g min}^{−1}\)

                                          \(= \frac{48}{32}\text{ mol min}^{−1}\)

                                           \(= 1.5 \text{ mol min}^{−1}\)

Rate of formation of \(H_2O\)

\( = \frac{1}{2}\frac{d[H_2O]}{dt} = \frac{d[O_2]}{dt}\)

\( = \frac{d[H_2O]}{dt} = \frac{2d[O_2]}{dt}\)

\( = \frac{d[H_2O]}{dt} = 2 × 1.5 \text{ mol min}^{−1}\)

\( = \frac{d[H_2O]}{dt} = 3 \text{ mol min}^{−1}\)