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Solve · Q4

Q.iv. What is the energy of activation of a reaction whose rate constant doubles when the temperature changes from 303 K to 313 K? (54.66 kJ/mol)

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Step 1. k doubles, so k2/k1=2k_2/k_1=2; T1 = 303 K, T2 = 313 K, R = 8.314 J/K/mol.

Step 2. log⁡102=Ea2.303×8.314×313−303303×313=Ea19.147×1094839\log_{10}2=\dfrac{E_a}{2.303\times8.314}\times\dfrac{313-303}{303\times313}=\dfrac{E_a}{19.147}\times\dfrac{10}{94839}.

Step 3. 0.30103=Ea×1019.147×94839=Ea×1018157300.30103=E_a\times\dfrac{10}{19.147\times94839}=E_a\times\dfrac{10}{1815730}. …

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