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Q.The rate constant for the first order decomposition of H2O2\text{H}_2\text{O}_2 is given by the following equation :
log⁡k=14.34−1.25×104 K/T\log k = 14.34 - 1.25\times10^{4}\,\text{K/T}
Calculate EaE_a for the reaction.
[Given : R=8.314 JK−1 mol−1R = 8.314\,\text{JK}^{-1}\,\text{mol}^{-1}]

Karnataka PUCKarnataka II PUC Board 2026Subjective· 3mImportance★★★★★
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Matching the given expression to the log form of the Arrhenius equation gives Ea=2.303 R×1.25×104≈239.3 kJ mol−1E_a = 2.303\,R \times 1.25\times10^4 \approx 239.3\,\text{kJ mol}^{-1}.

The logarithmic form of the Arrhenius equation is:

log⁡k=log⁡A−Ea2.303 RT\log k = \log A - \dfrac{E_a}{2.303\,RT}

Comparing with the given equation log⁡k=14.34−1.25×104 KT\log k = 14.34 - \dfrac{1.25\times10^4\,\text{K}}{T}, the coefficient of 1/T1/T gives:

Ea2.303 R=1.25×104 K\dfrac{E_a}{2.303\,R} = 1.25\times10^4\,\text{K}

Therefore:

Ea=2.303×R×1.25×104=2.303×8.314×1.25×104E_a = 2.303 \times R \times 1.25\times10^4 = 2.303 \times 8.314 \times 1.25\times10^4 …

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