The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be (R = 8.314 JK-1 mol-1 and log 2 = 0.301)

Question

The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be (R = 8.314 JK-1 mol-1 and log 2 = 0.301) (A) 53.6 kJ mol-1 (B) 48.6 kJ mol-1 (C) 58.5 kJ mol-1 (D) 60.5 kJ mol-1

Tardigrade Admin · Accepted Answer

From Arrhenius equation, log (k2/k1)=(-Ea/2.303 R)((1/T2)-(1/t1)) Given, (k2/k1)=2T2=310 K T1=300 K On putting values, ⇒ log 2=(-Ea/2.303×8.314)((1/310)-(1/300)) ⇒ Ea=53598.6 J = 53.6 kJ mol-1

Q. The rate of a reaction doubles when its temperature changes from $300 K$ to $310 K$ . Activation energy of such a reaction will be $(R = 8.314 J K^{- 1} m o l^{- 1}$ and $l o g 2 = 0.301)$

$53.6 k J m o l^{- 1}$

$48.6 k J m o l^{- 1}$

$58.5 k J m o l^{- 1}$

$60.5 k J m o l^{- 1}$

Q. The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be (R=8.314JK−1mol−1 and log2=0.301)

53.6kJmol−1

48.6kJmol−1

58.5kJmol−1

60.5kJmol−1

From Arrhenius equation, logk1​k2​​=2.303R−Ea​​(T2​1​−t1​1​) Given, k1​k2​​=2T2​=310K T1​=300K On putting values, ⇒log2=2.303×8.314−Ea​​(3101​−3001​) ⇒Ea​=53598.6J=53.6kJmol−1

Q. The rate of a reaction doubles when its temperature changes from $300 K$ to $310 K$ . Activation energy of such a reaction will be $(R = 8.314 J K^{- 1} m o l^{- 1}$ and $l o g 2 = 0.301)$

$53.6 k J m o l^{- 1}$

$48.6 k J m o l^{- 1}$

$58.5 k J m o l^{- 1}$

$60.5 k J m o l^{- 1}$