The rate constant of the first-order reaction may be described by following equation:

log K = $14.34 - \frac{1.25 \times 10^{4}}{T}$ , calculate energy of activation.

239.34kJ

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239.34 kcal

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239.34J

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239.34 cal.

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Solution

The correct option is A 239.34kJ
As we know that : $l n K = l n A - \frac{E_{a}}{R T}$ -------(1)
We have given $l o g K = 14.34 - \frac{1.25 \times 10^{4}}{T}$ -----(2)

Comparing (1) and (2) we will come to know that $\frac{E_{a}}{2.303 R} = 1.25 \times 10^{4}$

So $E_{a} = 2.303 \times 1.25 \times 10^{4} \times R$
$E_{a} = 2.303 \times 1.25 \times 10^{4} \times 8.314$
$E_{a} = 239.34 k J$

Option A is correct.

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The rate constant for the first order decomposition of $H_{2} O_{2}$ is given by the following equation, $l o g k = 14.34 - 1.25 \times 10^{4} K / T$ . Calculate $E_{a}$ for this reaction and at what temperature will its half-life period be 256 minutes?

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The rate constant of the first-order reaction may be described by following equation:log K = 14.34−1.25×104T, calculate energy of activation.

The correct option is A 239.34kJAs we know that : lnK=ln A−EaRT-------(1)We have given logK=14.34−1.25×104T-----(2)Comparing (1) and (2) we will come to know that Ea2.303R=1.25×104So Ea=2.303×1.25×104×REa=2.303×1.25×104×8.314Ea=239.34kJOption A is correct.

The rate constant of the first-order reaction may be described by following equation:

log K = $14.34 - \frac{1.25 \times 10^{4}}{T}$ , calculate energy of activation.