The rate of a reaction triples when temperature changes from $20^{o} C$ to $50^{o} C$ . Calculate energy of activation for the reaction. $(R = 8.314 J K^{- 1} {m o l}^{- 1})$

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Solution

The Arrhenius equation is,

${log}_{10} \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R \times 2.303} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

Given: $\frac{k_{2}}{k_{1}} = 3$ ; $R = 8.314$ $J$ $K^{- 1}$ ${m o l}^{- 1}$ ; $T_{1} = 20 + 273 = 293$ $K$
and $T_{2} = 50 + 273 = 323$ $K$

Substituting the given values in the Arrhenius equation,
${log}_{10} 3 = \frac{E_{a}}{8.314 \times 2.303} [\frac{323 - 293}{323 \times 293}]$

$E_{a} = \frac{2.303 \times 8.314 \times 323 \times 293 \times 0.477}{30}$

$= 28811.8$ $J$ ${m o l}^{- 1}$ $= 28.8118$ $k J$ ${m o l}^{- 1}$

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Similar questions

20^{o} C

50^{o} C

(R = 8.314 {JK}^{- 1} {mol}^{- 1}, l o g_{10} 3 = 0.477) .

20^{o} C t o 50^{o} C .

R = 8.314 J K^{- 1} m o l^{- 1}

27^{o} C

37^{o} C

[R = 8.314 J K^{- 1} m o l e]

300 K

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(R = 8.314 J K^{- 1} {m o l}^{- 1} a n d log 2 = 0.301)

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J . m o l^{- 1}

(l o g 3 = 0.4771, R = 8.314 K^{- 1} m o l^{- 1})

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