Question:

Consider the two different first-order reactions given below: A + BC (Reaction 1)PQ (Reaction 2)\text{A + B} \rightarrow \text{C (Reaction 1)} \\\text{P} \rightarrow \text{Q (Reaction 2)}The ratio of the half-life of Reaction 1 : Reaction 2 is 5:25 : 2. If t1t_1 and t2t_2 represent the time taken to complete 23rd\frac{2}{3}^\text{rd} and 45th\frac{4}{5}^\text{th} of Reaction 1 and Reaction 2, respectively, then the value of the ratio t1:t2t_1 : t_2 is ____ ×101\times 10^{-1} (nearest integer).[Given: log10(3)=0.477\log_{10}(3) = 0.477 and log10(5)=0.699\log_{10}(5) = 0.699]

Updated On: Mar 21, 2025
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Correct Answer: 17

Solution and Explanation

For first order reactions:

K1t1=ln(1123)=ln3 K_1 t_1 = \ln\left(\frac{1}{1 - \frac{2}{3}}\right) = \ln 3

K2t2=ln(1145)=ln5 K_2 t_2 = \ln\left(\frac{1}{1 - \frac{4}{5}}\right) = \ln 5

K1t1=5andK2t2=2 \therefore K_1 t_1 = 5 \quad \text{and} \quad K_2 t_2 = 2

K1K2=ln3ln5 \frac{K_1}{K_2} = \frac{\ln 3}{\ln 5}

t1t2=0.4770.699×5=1.7×101 \frac{t_1}{t_2} = \frac{0.477}{0.699} \times 5 = 1.7 \times 10^{-1}

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