For a decomposition reaction the values of the rate constant k at two different temperatures are given below: k$_1$ = 2.15 × 10$^{-8}$ L mol$^{-1}$ s$^{-1}$ at 650 K k$_2$ = 2.39 × 10$^{-7}$ L mol$^{-1}$ s$^{-1}$ at 700 K Calculate the value of the activation energy for this reaction. (R = 8.314 J K$^{-1}$ mol$^{-1}$)

CBSE Class 12 · 2009 · All India · Chemical KineticsShort Answer

For a decomposition reaction the values of the rate constant k at two different temperatures are given below:
k

_{1}

= 2.15 × 10

^{- 8}

L mol

^{- 1}

^{- 1}

at 650 K
k

_{2}

= 2.39 × 10

^{- 7}

L mol

^{- 1}

^{- 1}

at 700 K
Calculate the value of the activation energy for this reaction. (R = 8.314 J K

^{- 1}

mol

^{- 1}

)

Answer: E_a ≈ 182 kJ mol⁻¹

lo g \frac{k _{2}}{k _{1}} = lo g \frac{2.39 \times 1 0 ^{- 7}}{2.15 \times 1 0 ^{- 8}} = lo g 11.12 = 1.046

lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})

1.046 = \frac{E _{a}}{19.147} \times \frac{50}{650 \times 700}

E_{a} = \frac{1.046 \times 19.147 \times 455000}{50}

E_{a} = 1.82 \times 1 0^{5} J mo l^{- 1} = 182 kJ mo l^{- 1}

CBSE Class 12 Board · 2009 · All India · Subjective