By what factor would the specific rate of a first order reaction increase- if the temperature is increased from 27 - C to 47 - C- The activation energy for the reaction is 55-776 kJmol-1- Take ln2 - 0-7- R gas const -8-3 J-mol-K-1

From Arrhenius equation, ln k_2/k_1=E_a/R ( 1/T_1 1/T_2 ) ⇒ ln k_2/k_1=55776/8.3 ( 1/300 1/320 )=1.4 ⇒ ln k_2/k_1=2 × 0.7 = 2 ln 2=ln 4 ⇒k_2/k_1=4 ...

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Byju's Answer

Standard XII

Chemistry

Rate Constant

By what facto...

Question

By what factor would the specific rate of a first order reaction increase, if the temperature is increased from $27^{\circ} C$ to $47^{\circ} C$ . The activation energy for the reaction is $55.776 k J m o l^{- 1}$ .
(Take $l n 2 = 0.7, R (g a s c o n s t) = 8.3 J / (m o l . K)^{- 1})$

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Solution

From Arrhenius equation,
$l n \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$
$\Rightarrow l n \frac{k_{2}}{k_{1}} = \frac{55776}{8.3} (\frac{1}{300} - \frac{1}{320}) = 1.4$
$\Rightarrow l n \frac{k_{2}}{k_{1}} = 2 \times 0.7 = 2 l n 2 = l n 4$ $\Rightarrow \frac{k_{2}}{k_{1}} = 4$

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By what factor would the specific rate of a first order reaction increase, if the temperature is increased from 27 ∘C to 47 ∘C. The activation energy for the reaction is 55.776 kJmol−1. (Take ln2=0.7, R(gas const)=8.3 J/(mol.K)−1)

From Arrhenius equation, lnk2k1=EaR(1T1−1T2) ⇒lnk2k1=557768.3(1300−1320)=1.4 ⇒lnk2k1=2×0.7=2 ln 2=ln 4 ⇒k2k1=4

By what factor would the specific rate of a first order reaction increase, if the temperature is increased from $27^{\circ} C$ to $47^{\circ} C$ . The activation energy for the reaction is $55.776 k J m o l^{- 1}$ .
(Take $l n 2 = 0.7, R (g a s c o n s t) = 8.3 J / (m o l . K)^{- 1})$