Question

Activation energy ($E_a$) and rate constants (k₁ and k₂) of a chemical reaction at two different temperatures (T₁ and T₂) are related by :
(a) $ln \frac{k_2}{k_1} = - \frac{E_a}{R} \bigg \lgroup \frac{1}{T_1} - \frac{1}{T_2} \bigg \rgroup$
(b) $ln \frac{k_2}{k_1} = - \frac{E_a}{R} \bigg \lgroup \frac{1}{T_2} - \frac{1}{T_1} \bigg \rgroup$
(c) $ln \frac{k_2}{k_1} = - \frac{E_a}{R} \bigg \lgroup \frac{1}{T_2} + \frac{1}{T_1} \bigg \rgroup$
(d) $ln \frac{k_2}{k_1} = \frac{E_a}{R} \bigg \lgroup \frac{1}{T_1} - \frac{1}{T_2} \bigg \rgroup$

Answer 1

Here is your answer ⤵⤵⤵

(a) $ln \frac{k_2}{k_1} = - \frac{E_a}{R} \bigg \lgroup \frac{1}{T_1} - \frac{1}{T_2} \bigg \rgroup$

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Answer 2

$<b>$
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Activation energy ($E_a$) and rate constants (k₁ and k₂) of a chemical reaction at two different temperatures (T₁ and T₂) are related by :
(a) $ln \frac{k_2}{k_1} = - \frac{E_a}{R} \bigg \lgroup \frac{1}{T_1} - \frac{1}{T_2} \bigg \rgroup$
(b) $ln \frac{k_2}{k_1} = - \frac{E_a}{R} \bigg \lgroup \frac{1}{T_2} - \frac{1}{T_1} \bigg \rgroup$
(c) $ln \frac{k_2}{k_1} = - \frac{E_a}{R} \bigg \lgroup \frac{1}{T_2} + \frac{1}{T_1} \bigg \rgroup$
(d) $ln \frac{k_2}{k_1} = \frac{E_a}{R} \bigg \lgroup \frac{1}{T_1} - \frac{1}{T_2} \bigg \rgroup$


→ [B]


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