For a first order reaction, rate constant is given by lo g k = 14 − T 1.2 × 1 0 4 , then what will be the value of temperature if the rate constant is 100 sec − 1 ?

Question

For a first order reaction, rate constant is given by lo g k = 14 − T 1.2 × 1 0 4 ​ , then what will be the value of temperature if the rate constant is 100 sec − 1 ?

TestHub · Accepted Answer

Question Explanation: Find the value of temperature if the rate constant is $100 se c^{- 1}$ .

Concept: This question is based on the Arrhenius equation.

Solution:

$lo g K K lo g 100 2 \frac{1.2 \times 1 0 ^{4}}{T} T T T = 14 - \frac{1.2 \times 1 0 ^{4}}{T} = 100 se c^{- 1} = 14 - \frac{1.2 \times 1 0 ^{4}}{T} = 14 - \frac{1.2 \times 1 0 ^{4}}{T} = 12 = \frac{1.2 \times 1 0 ^{4}}{12} = 0.1 \times 1 0^{4} = 1000 K$

Final Answer: Option (C)

Chemistry - Chemical Kinetics Question with Solution | TestHub

Options:

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