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

2.8 k J

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12.5 k J m o l^{- 1}

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28.8 k J m o l^{- 1}

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61.2 k J m o l^{- 1}

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Solution

The correct option is C $28.8 k J m o l^{- 1}$
Expression for rate constant at two different temperature is given by
$l o g_{10} \frac{k_{2}}{k_{2}} = \frac{E_{a}}{R \times 2.303} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

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 equation,
$l o g_{10} 3 = \frac{E_{a}}{8.314 \times 2.303} [\frac{323 - 293}{323 \times 293}]$
$E_{a} = \frac{2.303 \times 8.314 \times 323 \times 293 \times 0.477}{30}$
$= 28811.8 J m o l^{- 1}$
$= 28.8118 k J m o l^{- 1}$
Hence, (c) is correct.

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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

50^{o} C

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

300 K

310 K

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

log 2 = 0.301]

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

l o g_{10} 2 = 0.301)

30^{0} C

50^{0} C

J m o l^{- 1} K^{- 1}

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