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ChemistryInteger+4 / -1Question 74 of 75

JEE Main 2026 · 2 Apr · Shift 1, Chemistry question 74

For reaction A→P\ce{A -> P}, rate constant k=1.5×103k=1.5\times10^3 s−1^{-1} at 27 °C.
If activation energy for the above reaction is 60 kJ mol−1^{-1}, then the temperature (in °C) at which rate constant, k=4.5×103k=4.5\times10^3 s−1^{-1} is ________. (Nearest integer)
Given : log⁡2=0.30\log2=0.30, log⁡3=0.48\log3=0.48, R=8.3R=8.3 J K−1^{-1} mol−1^{-1}, ln⁡10=2.3\ln10=2.3

Answer and worked solution

Official answer: 41

Arrhenius equation.
ln⁡k2k1=EaR(1T1−1T2),ln⁡3=0.48×2.3=1.104.\ln\frac{k_2}{k_1}=\frac{E_a}R\left(\frac1{T_1}-\frac1{T_2}\right),\qquad\ln3=0.48\times2.3=1.104.

Solve for T2T_2. 1T1−1T2=1.104×8.360000=1.527×10−4\frac1{T_1}-\frac1{T_2}=\frac{1.104\times8.3}{60000}=1.527\times10^{-4} K⁻¹, so 1T2=1300−1.527×10−4=3.181×10−3\frac1{T_2}=\frac1{300}-1.527\times10^{-4}=3.181\times10^{-3} and T2=314.4T_2=314.4 K.

In °C. 314.4−273≈41314.4-273\approx41 °C.

Answer: 41

Topics:Arrhenius equation
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