The specific rate for a reaction increases by a factor of 4- if the temperature is changed from 27 - C to 52 - C- Find the activation energy for the reaction in kJ mol-1 Take ln 2 - 0-7- R gas const - 253 J-mol K Round off the answer upto two decimal places

According to Arrhenius equation, k=Ae^ E_a/RT ⇒lnk=lnA E_aRT ln k_2k_1 = E_aR (1T_1 1T_2) ⇒ ln 4 = E_a253 (1300 1325) ⇒ 2 × 0.7 = 3× E_a25(25300 × 325) ⇒ E_ ...

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Standard XIII

Chemistry

Arrhenius Equation

The specific ...

Question

The specific rate for a reaction increases by a factor of 4, if the temperature is changed from $27^{\circ} C$ to $52^{\circ} C$ . Find the activation energy for the reaction in $k J m o l^{- 1}$ (Take $l n 2 = 0.7$ , R (gas const) = $\frac{25}{3} J / m o l K$ ) (Round off the answer upto two decimal places)

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Solution

According to Arrhenius equation,
$k = A e^{\frac{- E_{a}}{R T}}$
$\Rightarrow$ $l n k = l n A - \frac{E_{a}}{R T}$
$l n \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$
$\Rightarrow l n 4 = \frac{E_{a}}{\frac{25}{3}} (\frac{1}{300} - \frac{1}{325})$
$\Rightarrow 2 \times 0.7 = \frac{3 \times E_{a}}{25} (\frac{25}{300 \times 325}) \Rightarrow E_{a} = 45.5 \times 10^{3} J / m o l = 45.50 k J / m o l$

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The specific rate for a reaction increases by a factor of 4, if the temperature is changed from 27 ∘C to 52 ∘C. Find the activation energy for the reaction in kJ mol−1 (Take ln2=0.7, R (gas const) = 253 J/mol K) (Round off the answer upto two decimal places)

According to Arrhenius equation, k=Ae−EaRT ⇒lnk=lnA−EaRT lnk2k1=EaR(1T1−1T2) ⇒ln 4=Ea253(1300−1325) ⇒2×0.7=3×Ea25(25300×325)⇒Ea=45.5×103 J/mol=45.50 kJ/mol