For the reaction, a A + bB arrow cC + dD , the plot of log k vs (1/ T ) is given below: 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

Question

For the reaction, a A + bB arrow cC + dD , the plot of log k vs (1/ T ) is given below: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/0f708b79a4587619f0e0b50d6c421afd-.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 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

log K = log A -( Ea /2.303 RT ) Slope =( Ea /2.303 R )=10,000 log (( K 2/ K 1))=( Ea /2.303 R )((1/ T 1)-(1/ T 2)) log ((10-4/10-5))=10,000[(1/500)-(1/ T 2)] T 2=526.31 ≃ 526 K

Q. For the reaction, aA+bB→cC+dD, the plot of log k vs T1​ is given below : The temperature at which the rate constant of the reaction is 10−4s−1 is K. (Rounded-off to the nearest integer) [Given : The rate constant of the reaction is 10−5s−1 at 500K

logK=logA−2.303RTEa​ Slope =2.303REa​=10,000 log(K1​K2​​)=2.303REa​(T1​1​−T2​1​) log(10−510−4​)=10,000[5001​−T2​1​] T2​=526.31≃526K

$lo g K = lo g A - \frac{E a}{2.303 RT}$

Slope $= \frac{E a}{2.303 R} = 10, 000$

$lo g (\frac{K _{2}}{K _{1}}) = \frac{E a}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

$lo g (\frac{1 0 ^{- 4}}{1 0 ^{- 5}}) = 10, 000 [\frac{1}{500} - \frac{1}{T _{2}}]$

$T_{2} = 526.31 ≃ 526 K$