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JEE Main 2014 · 12 Apr · Shift 1, Chemistry question 35

The rate coefficient (k) for a particular reactions is 1.3×10−41.3\times10^{-4} M−1^{-1} s−1^{-1} at 100°C, and 1.3×10−31.3\times10^{-3} M−1^{-1} s−1^{-1} at 150°C. What is the energy of activation (EAE_A) (in kJ) for this reaction ? (R = molar gas constant = 8.314 JK−1^{-1} mol−1^{-1})

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Answer and worked solution

Official answer: B

ln⁡10=EAR(1373−1423)=EA8.314×3.17×10−4.\ln10=\frac{E_A}R\left(\frac1{373}-\frac1{423}\right)=\frac{E_A}{8.314}\times3.17\times10^{-4}.
EA=2.303×8.3143.17×10−4≈6.0×104 J≈60 kJ.E_A=\frac{2.303\times8.314}{3.17\times10^{-4}}\approx6.0\times10^4\text{ J}\approx60\text{ kJ}.

Topics:Arrhenius equation
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