JEE MAIN - Chemistry (2021 - 31st August Evening Shift - No. 18)
For the reaction A $$\to$$ B, the rate constant k(in s$$-$$1) is given by
$${\log _{10}}k = 20.35 - {{(2.47 \times {{10}^3})} \over T}$$
The energy of activation in kJ mol$$-$$1 is ____________. (Nearest integer) [Given : R = 8.314 J K$$-$$1 mol$$-$$1]
$${\log _{10}}k = 20.35 - {{(2.47 \times {{10}^3})} \over T}$$
The energy of activation in kJ mol$$-$$1 is ____________. (Nearest integer) [Given : R = 8.314 J K$$-$$1 mol$$-$$1]
Answer
47
Explanation
Given,
$$\log K = 20.35 - {{2.47 \times {{10}^3}} \over T}$$
We know
$$\log K = \log A - {{{E_a}} \over {2.303RT}}$$
$$ \Rightarrow {{{E_a}} \over {2.303RT}} = 2.47 \times {10^3}$$
$${E_a} = 2.47 \times {10^3} \times 2.303 \times {{8.314} \over {1000}}$$ KJ/mole
= 47.29 = 47 (Nearest integer)
$$\log K = 20.35 - {{2.47 \times {{10}^3}} \over T}$$
We know
$$\log K = \log A - {{{E_a}} \over {2.303RT}}$$
$$ \Rightarrow {{{E_a}} \over {2.303RT}} = 2.47 \times {10^3}$$
$${E_a} = 2.47 \times {10^3} \times 2.303 \times {{8.314} \over {1000}}$$ KJ/mole
= 47.29 = 47 (Nearest integer)
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