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NEET CHEMISTRYMedium

The plot of ln⁡k\ln k vs 1/T1/T for the following reaction, 2N2O5(g)→4NO2(g)+O2(g)2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g) gives a straight line with the slope of the line equal to −1.0×104 K-1.0 \times 10^4 \text{ K}. What is the activation energy for the reaction in J mol–1\text{J mol}^{–1}? (Given: R=8.3 J K–1 mol–1R = 8.3 \text{ J K}^{–1} \text{ mol}^{–1})

A

4.0×1024.0 \times 10^2

B

4.0×10−24.0 \times 10^{-2}

C

8.3×10−48.3 \times 10^{-4}

D

8.3×1048.3 \times 10^4

Step-by-Step Solution

According to the Arrhenius equation, k=Ae−Ea/RTk = A e^{-E_a/RT}. Taking the natural logarithm on both sides yields: ln⁡k=−EaR(1T)+ln⁡A\ln k = -\frac{E_a}{R} \left(\frac{1}{T}\right) + \ln A Comparing this with the equation of a straight line, y=mx+cy = mx + c, the plot of ln⁡k\ln k versus 1/T1/T gives a straight line with slope m=−EaRm = -\frac{E_a}{R}. Given slope =−1.0×104 K= -1.0 \times 10^4 \text{ K} −EaR=−1.0×104 K-\frac{E_a}{R} = -1.0 \times 10^4 \text{ K} Ea=1.0×104×RE_a = 1.0 \times 10^4 \times R Ea=1.0×104 K×8.3 J K−1 mol−1=8.3×104 J mol−1E_a = 1.0 \times 10^4 \text{ K} \times 8.3 \text{ J K}^{-1} \text{ mol}^{-1} = 8.3 \times 10^4 \text{ J mol}^{-1}.

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