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JEE Main 20262 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2026 · 2 April · Morning Shift · Chemistry · Chemical Kinetics

For reaction A arrow P , rate constant k = 1.5 × 10^3 s ^ -1 at 27°...

For reaction A→PA \rightarrow P, rate constant k=1.5×103k = 1.5 \times 10^3 s−1^{-1} at 27°27°C. If activation energy for the above reaction is 6060 kJ mol−1^{-1}, then the temperature (in °°C) at which rate constant, k=4.5×103k = 4.5 \times 10^3 s−1^{-1} is _______. (Nearest integer)
Given : log⁡2=0.30\log 2 = 0.30, log⁡3=0.48\log 3 = 0.48, R=8.3R = 8.3 J K−1^{-1} mol−1^{-1}, ln⁡10=2.3\ln 10 = 2.3

Correct answer

0

Step-by-step solution

Using the Arrhenius equation:

log⁡(k2k1)=Ea2.3R(1T1−1T2)\log \left( \dfrac{k_2}{k_1} \right) = \dfrac{E_a}{2.3 R} \left( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right)

Given:
k1=1.5×103 s−1k_1 = 1.5 \times 10^3 \text{ s}^{-1}
k2=4.5×103 s−1k_2 = 4.5 \times 10^3 \text{ s}^{-1}
T1=27∘C=300 KT_1 = 27^\circ\text{C} = 300 \text{ K}
Ea=60 kJ mol−1=60000 J mol−1E_a = 60 \text{ kJ mol}^{-1} = 60000 \text{ J mol}^{-1}
R=8.3 J K−1 mol−1R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}

Substituting the values into the equation:

log⁡(4.5×1031.5×103)=600002.3×8.3(1300−1T2)\log \left( \dfrac{4.5 \times 10^3}{1.5 \times 10^3} \right) = \dfrac{60000}{2.3 \times 8.3} \left( \dfrac{1}{300} - \dfrac{1}{T_2} \right)

log⁡3=6000019.09(1300−1T2)\log 3 = \dfrac{60000}{19.09} \left( \dfrac{1}{300} - \dfrac{1}{T_2} \right)

0.48=6000019.09(1300−1T2)0.48 = \dfrac{60000}{19.09} \left( \dfrac{1}{300} - \dfrac{1}{T_2} \right)

1300−1T2=0.48×19.0960000\dfrac{1}{300} - \dfrac{1}{T_2} = \dfrac{0.48 \times 19.09}{60000}

1300−1T2=9.163260000=1.5272×10−4\dfrac{1}{300} - \dfrac{1}{T_2} = \dfrac{9.1632}{60000} = 1.5272 \times 10^{-4}

1T2=1300−1.5272×10−4\dfrac{1}{T_2} = \dfrac{1}{300} - 1.5272 \times 10^{-4}

1T2=3.3333×10−3−0.1527×10−3=3.1806×10−3 K−1\dfrac{1}{T_2} = 3.3333 \times 10^{-3} - 0.1527 \times 10^{-3} = 3.1806 \times 10^{-3} \text{ K}^{-1}

T2=13.1806×10−3≈314.4 KT_2 = \dfrac{1}{3.1806 \times 10^{-3}} \approx 314.4 \text{ K}

Converting to Celsius:
T2=314.4−273=41.4∘CT_2 = 314.4 - 273 = 41.4^\circ\text{C}

The nearest integer is 4141.

Answer: 4141
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