CUET · CHEMISTRY · PYQ PAPER 2023
Decomposition of \(H _2 O _2\) follows a first order reaction.
In 50 min , the concentration of \(H _2 O _2\) decreases from 0.5 to 0.125 M.
When the concentration of \(H _2 O _2\) reaches 0.05 M.
the rate of formation of \(O _2\) will be (Given \(\log 2=0.3010\) ) :
- A \(1.34 \times 10^{-2} mol min ^{-1}\)
- B \(2.66 L min ^{-1}\) at STP
- C \(6.93 \times 10^{-4} mol min ^{-1}\)
- D \(6.93 \times 10^{-2} mol min ^{-1}\)
Answer & Solution
Correct Answer
(C) \(6.93 \times 10^{-4} mol min ^{-1}\)
Step-by-step Solution
Detailed explanation
\( k = \frac{1}{t} \ln \frac{[H_2O_2]_0}{[H_2O_2]_t} = \frac{1}{50 \text{ min}} \ln \frac{0.5 \text{ M}}{0.125 \text{ M}} \) \( k = \frac{1}{50} \ln 4 = \frac{2 \ln 2}{50} = \frac{2 \times (2.303 \times 0.3010)}{50} = 0.02772 \text{ min}^{-1} \) From…
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