For a reaction, the activation energy E_a = 0 and the rate constant at 200 K is 1.6× 106 s-1. The rate constant at 400 K will be (Given R = 8.314 J K-1mol-1):

Correct answer: (B) 1.6× 106 s-1. Answer: (B) 1.6× 106 s-1 Solution: Using the two-temperature form of the Arrhenius equation: _2/k_1 = E_a/2.303,R (1/T_1 - 1/T_2 ) Since E_a = 0, the right-hand side is 0, so _2/k_1 = 0, giving k_2 = k_1. With zero activation energy the rate constant is independent of temperature, so the rate constant at 400 K equals that at 200 K = 1.6× 106 s-1.

NEET 2019 Odisha · ChemistryArrheniusPrevious Year Question

For a reaction, the activation energy $E_{a} = 0$ and the rate constant at $200$ K is $1.6 \times 1 0^{6} s^{- 1}$ . The rate constant at $400$ K will be (Given $R = 8.314 J K^{- 1} mol^{- 1}$ ):

Answer: (B) $1.6 \times 1 0^{6} s^{- 1}$ . \textbf{Answer:} (B) $1.6 \times 1 0^{6} s^{- 1}$ \textbf{Solution:} Using the two-temperature form of the Arrhenius equation: $lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$ Since $E_{a} = 0$ , the right-hand side is $0$ , so $lo g \frac{k _{2}}{k _{1}} = 0$ , giving $k_{2} = k_{1}$ .

A. $3.2 \times 1 0^{4} s^{- 1}$
B. $1.6 \times 1 0^{6} s^{- 1}$ ✓
C. $1.6 \times 1 0^{3} s^{- 1}$
D. $3.2 \times 1 0^{6} s^{- 1}$

Correct Answer

(B) $1.6 \times 1 0^{6} s^{- 1}$

Solution & Explanation

\textbf{Answer:} (B) $1.6 \times 1 0^{6} s^{- 1}$ \textbf{Solution:} Using the two-temperature form of the Arrhenius equation: $lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$ Since $E_{a} = 0$ , the right-hand side is $0$ , so $lo g \frac{k _{2}}{k _{1}} = 0$ , giving $k_{2} = k_{1}$ . With zero activation energy the rate constant is independent of temperature, so the rate constant at $400$ K equals that at $200$ K $= 1.6 \times 1 0^{6} s^{- 1}$ .

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For a reaction, the activation energy Ea​=0 and the rate constant at 200 K is 1.6×106 s−1. The rate constant at 400 K will be (Given R=8.314 J K−1mol−1):

Solution & Explanation

More NEET PYQ solutions

For a reaction, the activation energy $E_{a} = 0$ and the rate constant at $200$ K is $1.6 \times 1 0^{6} s^{- 1}$ . The rate constant at $400$ K will be (Given $R = 8.314 J K^{- 1} mol^{- 1}$ ):