The activation energy for a reaction which doubles the rate when the temperature is raised from 298 K to 308 K is

Question

The activation energy for a reaction which doubles the rate when the temperature is raised from 298 K to 308 K is (A) 59.2 kJ mol-1 (B) 39.2 kJ mol-1 (C) 52.9 kJ mol-1 (D) 29.5 kJ mol-1

Tardigrade Admin · Accepted Answer

Activation energy can be calculated from the equation.

\frac{l o g K _{2}}{l o g K _{1}} = \frac{- E _{a}}{2.303 R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})

Given

\frac{l o g K _{2}}{l o g K _{1}} = 2 T_{2} = 308; T_{1} = 298

∴ l o g 2 = \frac{- E _{a}}{2.303 \times 8.314} (\frac{1}{308} - \frac{1}{298})

E_{a} = 52.9 k J m o l^{- 1}

Q. The activation energy for a reaction which doubles the rate when the temperature is raised from 298 K to 308 K is

$59.2 k J m o l^{- 1}$

$39.2 k J m o l^{- 1}$

$52.9 k J m o l^{- 1}$

$29.5 k J m o l^{- 1}$

Q. The activation energy for a reaction which doubles the rate when the temperature is raised from 298 K to 308 K is

59.2kJmol−1

39.2kJmol−1

52.9kJmol−1

29.5kJmol−1

Activation energy can be calculated from the equation. logK1​logK2​​=2.303R−Ea​​(T2​1​−T1​1​) Given logK1​logK2​​=2T2​=308;T1​=298 ∴log2=2.303×8.314−Ea​​(3081​−2981​) Ea​=52.9kJmol−1

$59.2 k J m o l^{- 1}$

$39.2 k J m o l^{- 1}$

$52.9 k J m o l^{- 1}$

$29.5 k J m o l^{- 1}$