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CBSE Class 12 Chemistry 2023 Delhi Set 1 Solved Paper

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Question : 26 of 35
Marks: +1, -0
SECTION - C
(a) For the reaction
2N2O5(g)→4NO2(g)+O2(g) at 318K2 \mathrm{N}_2 \mathrm{O}_{5(g)} \rightarrow 4 \mathrm{NO}_{2(g)} + \mathrm{O}_{2(g)} \text{ at } 318 \mathrm{K}
Calculate the rate of reaction if rate of disappearance of N2O5(g)\mathrm{N}_2\mathrm{O}_{5(g)} is 1.4×10−3ms−11.4 \times 10^{-3} \mathrm{m} \mathrm{s}^{-1}.
(b) For a first order reaction derive the relationshipt99%=2t90%t_{99\%} = 2 t_{90\%}
Solution:     👈: Video Solution
(a) 2N2O5→4NO2+O22 \mathrm{N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2 + \mathrm{O}_2
−12d[N2O5]dt=+14d[NO2]dt=+d[O2]dt\frac{-\frac{1}{2} d[\mathrm{N}_2\mathrm{O}_5]}{dt} = \frac{+\frac{1}{4} d[\mathrm{NO}_2]}{dt} = \frac{+ d[\mathrm{O}_2]}{dt}
=2×14×1.4×10−3= 2 \times \frac{1}{4} \times 1.4 \times 10^{-3} [Rate of disappearance =1.4×10−3 ms−1]= 1.4 \times 10^{-3} \text{ ms}^{-1} ]
=0.7×10−3ms−1= 0.7 \times 10^{-3} \mathrm{ms}^{-1}
(b) For a first order of reaction
t=2.303Klog⁡aa−xt = \frac{2.303}{K} \log \frac{a}{a-x}
t99%=2.303Klog⁡1001t_{99\%} = \frac{2.303}{K} \log \frac{100}{1}
=2.303Klog⁡100= \frac{2.303}{K} \log 100
=2.303×2K=4.606K= \frac{2.303 \times 2}{K} = \frac{4.606}{K}
and t90%=2.303Klog⁡aa−xt_{90\%} = \frac{2.303}{K} \log \frac{a}{a-x}
=2.303Klog⁡10=2.303K= \frac{2.303}{K} \log 10 = \frac{2.303}{K}
t99%t90%=2\frac{t_{99\%}}{t_{90\%}} = 2
t99%=2t90%t_{99\%} = 2 t_{90\%}.
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