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Chemical Kinetics and Nuclear Chemistry question

2016 · Shift 0 · Q1
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Chemical Kinetics and Nuclear Chemistry question

2016 · Shift 0 · Q1

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
Decomposition of H2O2H_2O_2H2​O2​ follows a first order reaction. In fifty minutes the concentration of H2O2H_2O_2H2​O2​ decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2H_2O_2H2​O2​ reaches 0.05 M, the rate of formation of O2O_2O2​ will be :
  1. A
    6.93 ×\times× 10-4 mol min-1
  2. B
    6.96 ×\times× 10-2 mol min-1
  3. C
    1.34 ×\times× 10-2 mol min-1
  4. D
    2.66 L min–1 at STP
View written solutionFree

Correct answer: A

  1. Write the reaction and rate relation

The decomposition is: 2H2O2→2H2O+O22H_2O_2 \rightarrow 2H_2O + O_22H2​O2​→2H2​O+O2​

Since it is a first order reaction in H2O2H_2O_2H2​O2​, k=2.303tlog⁡[H2O2]0[H2O2]tk = \frac{2.303}{t}\log\frac{[H_2O_2]_0}{[H_2O_2]_t}k=t2.303​log[H2​O2​]t​[H2​O2​]0​​

Given:

  • [H2O2]0=0.5 M[H_2O_2]_0 = 0.5\,\text{M}[H2​O2​]0​=0.5M
  • [H2O2]t=0.125 M[H_2O_2]_t = 0.125\,\text{M}[H2​O2​]t​=0.125M
  • t=50 mint = 50\,\text{min}t=50min

So, k=2.30350log⁡0.50.125k = \frac{2.303}{50}\log\frac{0.5}{0.125}k=502.303​log0.1250.5​ =2.30350log⁡4= \frac{2.303}{50}\log 4=502.303​log4

Using log⁡4=0.6021\log 4 = 0.6021log4=0.6021, k=2.303×0.602150k = \frac{2.303 \times 0.6021}{50}k=502.303×0.6021​ k≈1.38650k \approx \frac{1.386}{50}k≈501.386​ k≈2.772×10−2 min−1k \approx 2.772 \times 10^{-2}\,\text{min}^{-1}k≈2.772×10−2min−1

  1. Find the rate of disappearance of H2O2H_2O_2H2​O2​ when [H2O2]=0.05 [H_2O_2]=0.05\,[H2​O2​]=0.05M

For a first order reaction, −d[H2O2]dt=k[H2O2]-\frac{d[H_2O_2]}{dt} = k[H_2O_2]−dtd[H2​O2​]​=k[H2​O2​]

Thus, −d[H2O2]dt=(2.772×10−2)(0.05)-\frac{d[H_2O_2]}{dt} = (2.772 \times 10^{-2})(0.05)−dtd[H2​O2​]​=(2.772×10−2)(0.05) =1.386×10−3 mol L−1min−1= 1.386 \times 10^{-3}\,\text{mol L}^{-1}\text{min}^{-1}=1.386×10−3mol L−1min−1

  1. Relate this to rate of formation of O2O_2O2​

From 2H2O2→O2+2H2O2H_2O_2 \rightarrow O_2 + 2H_2O2H2​O2​→O2​+2H2​O

For every 2 moles of H2O2H_2O_2H2​O2​ decomposed, 1 mole of O2O_2O2​ is formed.

Hence, d[O2]dt=12(−d[H2O2]dt)\frac{d[O_2]}{dt} = \frac{1}{2}\left(-\frac{d[H_2O_2]}{dt}\right)dtd[O2​]​=21​(−dtd[H2​O2​]​)

So, d[O2]dt=12(1.386×10−3)\frac{d[O_2]}{dt} = \frac{1}{2}(1.386 \times 10^{-3})dtd[O2​]​=21​(1.386×10−3) =6.93×10−4 mol L−1min−1= 6.93 \times 10^{-4}\,\text{mol L}^{-1}\text{min}^{-1}=6.93×10−4mol L−1min−1

  1. Match with the options

This corresponds to: 6.93×10−4 mol min−1\boxed{6.93 \times 10^{-4}\,\text{mol min}^{-1}}6.93×10−4mol min−1​

So the correct option is A.

Note: Strictly, the unit from concentration data should be mol L−1min−1\text{mol L}^{-1}\text{min}^{-1}mol L−1min−1, but the intended option is clearly A.

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