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NEET CHEMISTRYMedium

The slope of Arrhenius plot (ln⁡k v/s 1T)\left( \ln k \text{ v/s } \frac{1}{T} \right) of first order reaction is −5×103K-5 \times 10^3 \text{K}. The value of EaE_a of the reaction is. Choose the correct option for your answer. [Given R=8.314 JK−1mol−1R = 8.314 \text{ JK}^{-1}\text{mol}^{-1}]

A

41.5 kJ mol−141.5 \text{ kJ mol}^{-1}

B

83.0 kJ mol−183.0 \text{ kJ mol}^{-1}

C

166 kJ mol−1166 \text{ kJ mol}^{-1}

D

−83 kJ mol−1-83 \text{ kJ mol}^{-1}

Step-by-Step Solution

Arrhenius equation: k=Ae−Ea/RTk = Ae^{-E_a/RT}. Taking natural log: ln⁡k=ln⁡A−EaR(1T)\ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} \right). The slope m=−EaRm = -\frac{E_a}{R}. Given m=−5×103Km = -5 \times 10^3 \text{K}. So, −5×103=−Ea8.314-5 \times 10^3 = -\frac{E_a}{8.314}. Ea=5×103×8.314=41570 J/mol=41.57 kJ/mol≈41.5 kJ/molE_a = 5 \times 10^3 \times 8.314 = 41570 \text{ J/mol} = 41.57 \text{ kJ/mol} \approx 41.5 \text{ kJ/mol}.

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