SATHEE: Chemical Kinetics - Result Question 71

Chemical Kinetics - Result Question 71

####70. A first order reaction, $A \to B$ , requires activation energy of $70 k J m o l^{- 1}$ . When a $20$ solution of $A$ was kept at $25^{\circ} C$ for $20 m i n, 25$ decomposition took place. What will be the percentage decomposition in the same time in a $30$ solution maintained at $40^{\circ} C$ ? Assume that activation energy remains constant in this range of temperature.

$(1993, 4 M)$

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Solution:

$\ln \frac{k (40^{\circ} C)}{k (25^{\circ} C)} = \frac{E_{a}}{R} (\frac{15}{298 \times 313})$

$\begin{aligned} = \frac{70 \times 1000}{8.314} \times \frac{15}{298 \times 313} = 1.35 & \Rightarrow \frac{k (40^{\circ} C)}{k (25^{\circ} C)} = 3.87 & Also k (25^{\circ} C) = \frac{1}{20} \ln \frac{100}{75} = \frac{1}{20} \ln \frac{4}{3} & \Rightarrow k (40^{\circ} C) = 3.87 \times k (25^{\circ} C) & = 3.87 \times \frac{1}{20} \ln \frac{4}{3} = 55.66 \times 10^{- 3} m i n^{- 1} & Now k (40^{\circ} C) \times 20 = \ln \frac{100}{100 - x} & \Rightarrow 55.66 \times 10^{- 3} \times 20 = \ln \frac{100}{100 - x} \Rightarrow x = 67 \end{aligned}$

Chemical Kinetics - Result Question 71

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Chemical Kinetics - Result Question 71

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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