JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
Decomposition of \(H_2O\) follows a first order reaction. In fifty minutes the concentration of \(H_2O_2\) decreases from \(0.5\) to \(0.125\,\min\) one such decomposition. When the concentration of \(H_2O_2\) reaches \(0.05\, M,\) the rate of formation of \(O_2\) will be :
- A \(2.66\,L\,min^{-1}\,\) at \(STP\)
- B \(1.34 \times 10-^{-2} \,mol\, min^{-1}\)
- C \(6.96 \times 10-^{-2} \,mol\, min^{-1}\)
- D \(6.93 \times 10-^{-2} \,mol\, min^{-1}\)
Answer & Solution
Correct Answer
(D) \(6.93 \times 10-^{-2} \,mol\, min^{-1}\)
Step-by-step Solution
Detailed explanation
\(H_{2} O_{2}(a q) \rightarrow H_{2} O(a q)+\frac{1}{2} O_{2}(g)\) For a first order reaction \(k=\frac{2.303}{t} \log \frac{a}{(a-x)}\) Given \(a=0.5,(a-x)=0.125,\,t=50\,min\) \(k=\frac{2.303}{50} \log \frac{0.5}{0.125}\) \(=2.78 \times 10^{-2} \,\mathrm{min}^{-1}\)…
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