The differential form of the Arrhenius equati | vedclass

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Question

(B) The Arrhenius equation is given by $K = A e^{-\frac{E_a}{RT}}$.Taking the natural logarithm on both sides,we get $\ln K = \ln A - \frac{E_a}{RT}$.

Vedclass · Accepted Answer

$\frac{d \ln K}{dT} = \frac{\Delta E^*}{RT^2}$

Vedclass · Answer

(B) The Arrhenius equation is given by $K = A e^{-\frac{E_a}{RT}}$.Taking the natural logarithm on both sides,we get $\ln K = \ln A - \frac{E_a}{RT}$.Differentiating both sides with respect to temperature $T$,we get $\frac{d \ln K}{dT} = 0 - \frac{E_a}{R} 	imes (-\frac{1}{T^2})$.Therefore,$\frac{d \ln K}{dT} = \frac{E_a}{RT^2}$.Comparing this with the given options,where $\Delta E^*$ represents the activation energy $E_a$,the correct expression is $\frac{d \ln K}{dT} = \frac{\Delta E^*}{RT^2}$.

$Case$	$E_a \ (kcal/mol)$	$Temp. \ Change \ (K)$
$I$	$40$	$200 - 210$
$II$	$80$	$200 - 210$
$III$	$40$	$300 - 310$
$IV$	$80$	$300 - 310$

The differential form of the Arrhenius equation is:

Similar Questions

What is the value of the slope in the graph of $\log_{10} K$ against $\frac{1}{T}$?

Which one of the following is true for an exothermic reaction $A \rightleftharpoons B$,if $E_f$ and $E_b$ are the activation energies of forward and backward reactions respectively?

In which of the following cases is the percentage increase in rate constant maximum?
$Case$ $E_a \ (kcal/mol)$ $Temp. \ Change \ (K)$
$I$ $40$ $200 - 210$
$II$ $80$ $200 - 210$
$III$ $40$ $300 - 310$
$IV$ $80$ $300 - 310$

According to the Arrhenius equation,in a graph of $\log k$ versus $\frac{1}{T}$,the slope will be . . . . . . .

$A$ catalyst lowers the activation energy for a certain reaction from $83.314 \, kJ \, mol^{-1}$ to $75 \, kJ \, mol^{-1}$ at $500 \, K$. What will be the rate of reaction as compared to the uncatalysed reaction? Assume other things being equal.

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