The slope of the Arrhenius plot ( k vs 1/T) of a first-order reaction is -5× 103 K. The value of E_a of the reaction is (Given R = 8.314 J K-1mol-1):

Correct answer: (C) 41.5 kJ mol-1. Answer: (C) 41.5 kJ mol-1 Solution: The logarithmic form of the Arrhenius equation is: k = A - E_a/RT A plot of k vs 1/T is a straight line of slope = - E_aR. Therefore: E_a = -(slope)× R = -(-5× 103)× 8.314 = 41570 J mol-1 41.5 kJ mol-1

NEET 2021 · ChemistryArrheniusPrevious Year Question

The slope of the Arrhenius plot ( $ln k$ vs $1/ T$ ) of a first-order reaction is $- 5 \times 1 0^{3}$ K. The value of $E_{a}$ of the reaction is (Given $R = 8.314 J K^{- 1} mol^{- 1}$ ):

Answer: (C) $41.5 kJ mol^{- 1}$ . \textbf{Answer:} (C) $41.5 kJ mol^{- 1}$ \textbf{Solution:} The logarithmic form of the Arrhenius equation is: $ln k = ln A - \frac{E _{a}}{R T}$ A plot of $ln k$ vs $1/ T$ is a straight line of slope $= - \frac{E _{a}}{R}$ .

A. $166 kJ mol^{- 1}$
B. $- 83 kJ mol^{- 1}$
C. $41.5 kJ mol^{- 1}$ ✓
D. $83.0 kJ mol^{- 1}$

Correct Answer

Solution & Explanation

\textbf{Answer:} (C) $41.5 kJ mol^{- 1}$ \textbf{Solution:} The logarithmic form of the Arrhenius equation is: $ln k = ln A - \frac{E _{a}}{R T}$ A plot of $ln k$ vs $1/ T$ is a straight line of slope $= - \frac{E _{a}}{R}$ . Therefore: $E_{a} = - (slope) \times R = - (- 5 \times 1 0^{3}) \times 8.314 = 41570 J mol^{- 1} \approx 41.5 kJ mol^{- 1}$

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The slope of the Arrhenius plot (lnk vs 1/T) of a first-order reaction is −5×103 K. The value of Ea​ of the reaction is (Given R=8.314 J K−1mol−1):

Solution & Explanation

More NEET PYQ solutions

The slope of the Arrhenius plot ( $ln k$ vs $1/ T$ ) of a first-order reaction is $- 5 \times 1 0^{3}$ K. The value of $E_{a}$ of the reaction is (Given $R = 8.314 J K^{- 1} mol^{- 1}$ ):