1. Tardigrade
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  3. Chemistry
  4. The activation energy of one of the reactions in a biochemical process is 532611 J mol -1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k 300= x × 10-3 k 310°. The value of x is [Given: ln 10=2.3 R =8.3 J K -1 mol -1 text ]

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Q. The activation energy of one of the reactions in a biochemical process is $532611 \,J\, mol ^{-1}$. When the temperature falls from $310 K$ to $300 K$, the change in rate constant observed is $k _{300}= x \times 10^{-3} k _{310^{\circ}}$. The value of $x $ is____ [Given: $\ln 10=2.3$
$R =8.3 \,J \,K ^{-1} mol ^{-1} \text { ] }$

JEE MainJEE Main 2022Chemical Kinetics

Solution:

$ln \left(\frac{ K _{2}}{ K _{1}}\right)=\frac{ E _{ a }}{ R }\left(\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right)$
$ln \left(\frac{ K _{2}}{ K _{1}}\right)=\frac{532611}{8.3} \times\left(\frac{10}{310 \times 300}\right)$
where $K _{2}$ is at $310 K \& K _{1}$ is at $300\, K$
$ln \left(\frac{ K _{2}}{ K _{1}}\right)=6.9$
$= 3 \times \ell n 10 $
$ln \frac{ K _{2}}{ K _{1}}= ln 10^{3} $
$K _{2}= K _{1} \times 10^{3}$
$K _{1}= K _{2} \times 10^{3}$
So, $K = 1$