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Question

(B) The temperature coefficient $(n)$ is defined as the ratio of rate constants at temperatures differing by $10^{\circ} C$: $n = \frac{k_{T+10}}{k_T}

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(B) The temperature coefficient $(n)$ is defined as the ratio of rate constants at temperatures differing by $10^{\circ} C$: $n = \frac{k_{T+10}}{k_T} = 2$. This means for every $10^{\circ} C$ rise in temperature,the rate of reaction doubles. The total rise in temperature is $\Delta T = 90^{\circ} C - 30^{\circ} C = 60^{\circ} C$. The number of $10^{\circ} C$ intervals is $x = \frac{60}{10} = 6$. The rate of reaction increases by a factor of $n^x = 2^6$. $2^6 = 2 	imes 2 	imes 2 	imes 2 	imes 2 	imes 2 = 64$. Therefore,the rate of reaction increases by $64$ times.

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

The temperature coefficient of a reaction is $2$. When the temperature is increased from $30^{\circ} C$ to $90^{\circ} C$,the rate of reaction is increased by: (in $times$)

Similar Questions

The rate constant for a chemical reaction taking place at $500 \ K$ is expressed as $K = A \ e^{-1000}$. The activation energy of the reaction is:

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

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Which of the following graphs of $\log \, K \rightarrow 1/T$ can be used to calculate the activation energy?

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