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CBSE Class 12 Chemistry 2020 Outside Delhi Set 1 Solved Paper

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Question : 29 of 37
Marks: +1, -0
The rate constant for the first order decomposition of N2O5\mathrm{N}_2\mathrm{O}_5 is given by the following equation : k=(2.5×1014 s−1)e(−25000 K)/Tk = (2.5 \times 10^{14} \, \mathrm{s}^{-1}) e^{(-25000 \, \mathrm{K}) / T}
Calculate EaE_a for this reaction and rate constant if its half-life period be 300 minutes.
Solution:  
  k=(2.5×1014 l−1)e(−25000 K)/T\; k = (2.5 \times 10^{14} \, \mathrm{l}^{-1}) e^{(-25000 \, \mathrm{K}) / T}
  t1/2=300   minutes   \; t_{1/2} = 300 \; \text{ minutes } \;
    EaR=25000 K\;\; \frac{E_a}{R} = 25000 \, \mathrm{K}
  Ea=25000×R×K\; E_a = 25000 \times R \times \mathrm{K}
  =25000×8.314 J K−1 mol−1×K\; = 25000 \times 8.314 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1} \times \mathrm{K}
  =207850 J mol−1=207.850 kJ mol−1\; = 207850 \, \mathrm{J} \, \mathrm{mol}^{-1} = 207.850 \, \mathrm{kJ} \, \mathrm{mol}^{-1}
  t1/2=0.693K⇒K=0.693300 min−1\; t_{1/2} = \frac{0.693}{K} \Rightarrow K = \frac{0.693}{300} \, \mathrm{min}^{-1}
  =0.231×10−2\; = 0.231 \times 10^{-2}
  =2.31×10−3 min−1\; = 2.31 \times 10^{-3} \, \mathrm{min}^{-1}
  \;
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