JEE MAIN - Chemistry (2021 - 25th February Morning Shift - No. 23)
For the reaction, aA + bB $$ \to $$ cC + dD, the plot of log k vs $${1 \over T}$$ is given below :
_25th_February_Morning_Shift_en_23_1.png)
The temperature at which the rate constant of the reaction is 10-4 s-1 is _________ K. (Rounded off to the nearest integer)
[Given : The rate constant of the reaction is 10-5 s-1 at 500 K.]
The temperature at which the rate constant of the reaction is 10-4 s-1 is _________ K. (Rounded off to the nearest integer)
[Given : The rate constant of the reaction is 10-5 s-1 at 500 K.]
Answer
526
Explanation
$${\log _{10}}K = {\log _{10}}A - {{{E_a}} \over {2.303RT}}$$
Slope $$ = {{{E_a}} \over {2.303R}} = - 10000$$
$${\log _{10}}{{{K_2}} \over {{K_1}}} = {{{E_a}} \over {2.303R}} \times \left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$$
$${\log _{10}}{{{{10}^{ - 4}}} \over {{{10}^{ - 5}}}} = 10000 \times \left[ {{1 \over {500}} - {1 \over T}} \right]$$
$$ \Rightarrow $$ $$1 = 10000 \times \left[ {{1 \over {500}} - {1 \over T}} \right]$$
$$ \Rightarrow $$ $${1 \over {10000}} = {1 \over {500}} - {1 \over T}$$
$$ \Rightarrow $$ $${1 \over T} = {1 \over {500}} - {1 \over {10000}}$$
$$ \Rightarrow $$$${1 \over T} = {{20 - 1} \over {10000}} = {{19} \over {10000}}$$
$$ \Rightarrow $$ $$T = {{10000} \over {19}} = 526$$ K
Slope $$ = {{{E_a}} \over {2.303R}} = - 10000$$
$${\log _{10}}{{{K_2}} \over {{K_1}}} = {{{E_a}} \over {2.303R}} \times \left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$$
$${\log _{10}}{{{{10}^{ - 4}}} \over {{{10}^{ - 5}}}} = 10000 \times \left[ {{1 \over {500}} - {1 \over T}} \right]$$
$$ \Rightarrow $$ $$1 = 10000 \times \left[ {{1 \over {500}} - {1 \over T}} \right]$$
$$ \Rightarrow $$ $${1 \over {10000}} = {1 \over {500}} - {1 \over T}$$
$$ \Rightarrow $$ $${1 \over T} = {1 \over {500}} - {1 \over {10000}}$$
$$ \Rightarrow $$$${1 \over T} = {{20 - 1} \over {10000}} = {{19} \over {10000}}$$
$$ \Rightarrow $$ $$T = {{10000} \over {19}} = 526$$ K
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