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:
Answer & explanation
Correct answer: option 4
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 \(H_2O\)
\( = \frac{1}{2}\frac{d[H_2O]}{dt} = \frac{d[O_2]}{dt}\)
\( = \frac{d[H_2O]}{dt} = \frac{2d[O_2]}{dt}\)
\( = \frac{d[H_2O]}{dt} = 2 × 1.5 \text{ mol min}^{−1}\)
\( = \frac{d[H_2O]}{dt} = 3 \text{ mol min}^{−1}\)