"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]"

Question

Accepted Answer

"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)"

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]

Solution

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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]

Solution

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More Chemical Kinetics questions

Drill 25 more like these. Every day. Free.

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]