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CBSE Class 12 Chemistry 2015 Solved Paper

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Question : 24 of 26
Marks: +1, -0
(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(Given log⁡2=0.3010,log⁡4=0.6021\log 2=0.3010, \log 4=0.6021 )
OR
(a) For a reaction A+B→PA + B \rightarrow P, the rate is given by Rate =k[A][B]2=k[A][B]^2
(i) How is the rate of reaction affected if the concentration of BB is doubled ?
(ii) What is the overall order of reaction if AA is present in large excess?
(b) A first order reaction takes 303 0 minutes for 50%5 0 \% completion. Calculate the time required for 90%9 0 \% completion of this reaction.
(log⁡2=0.3010)(\log 2=0.3010)
Solution:  
  k=  2.303tlog⁡[A]0[A]\;k=\;\frac{2.303}{t} \log \frac{[A]_0}{[A]}
  k=  2.30330log⁡0.600.30\;k=\;\frac{2.303}{30} \log \frac{0.60}{0.30}
  k=  2.30330×0.301=0.023  s−1\;k=\;\frac{2.303}{30} \times 0.301=0.023 \; \mathrm{s}^{-1}
  k=  2.30360log⁡0.600.15\;k=\;\frac{2.303}{60} \log \frac{0.60}{0.15}
  k=  2.30360×0.6021=0.023  s−1\;k=\;\frac{2.303}{60} \times 0.6021=0.023 \; \mathrm{s}^{-1}
(ii) As kk is constant in both the readings, hence it is a pseudo-first order reaction.
   Rate   =−Δ[R]Δt\; \text{ Rate } \; = -\frac{\Delta[R]}{\Delta t}
  =−[0.15−0.30]60−30\; = -\frac{[0.15-0.30]}{60-30}
  =0.005  mol  L−1  s−1\; = 0.005 \; \mathrm{mol} \; \mathrm{L}^{-1} \; \mathrm{s}^{-1}
OR
(a) (i) Rate will increase 4 times of the actual rate of reaction.
(ii) Second order reaction
(b) t1/2  =  0.693kt_{1/2} \; = \; \frac{0.693}{k}
30  min  =  0.693k30 \; \mathrm{min} \; = \; \frac{0.693}{k}
k  =0.0231  min−1k \; = 0.0231 \; \mathrm{min}^{-1}
t  =  2.303tlog⁡[A]0[A]t \; = \; \frac{2.303}{t} \log \frac{[A]_0}{[A]}
t  =  2.3030.0231  mint \; = \; \frac{2.303}{0.0231} \; \mathrm{min}
t  =99.7  mint \; = 99.7 \; \mathrm{min}
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