JEE Main 2026 — Chemistry Physical Chemistry

Using the Arrhenius equation:

$$\log \left( \dfrac{k_2}{k_1} \right) = \dfrac{E_a}{2.3 R} \left( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right)$$log(k1​k2​​)=2.3REa​​(T1​1​−T2​1​)

Given:

$$k_1 = 1.5 \times 10^3 \text{ s}^{-1}$$k1​=1.5×103 s−1

$$k_2 = 4.5 \times 10^3 \text{ s}^{-1}$$k2​=4.5×103 s−1

$$T_1 = 27^\circ\text{C} = 300 \text{ K}$$T1​=27∘C=300 K

$$E_a = 60 \text{ kJ mol}^{-1} = 60000 \text{ J mol}^{-1}$$Ea​=60 kJ mol−1=60000 J mol−1

$$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$$R=8.3 J K−1 mol−1

Substituting the values into the equation:

$$\log \left( \dfrac{4.5 \times 10^3}{1.5 \times 10^3} \right) = \dfrac{60000}{2.3 \times 8.3} \left( \dfrac{1}{300} - \dfrac{1}{T_2} \right)$$log(1.5×1034.5×103​)=2.3×8.360000​(3001​−T2​1​)

$$\log 3 = \dfrac{60000}{19.09} \left( \dfrac{1}{300} - \dfrac{1}{T_2} \right)$$log3=19.0960000​(3001​−T2​1​)

$$0.48 = \dfrac{60000}{19.09} \left( \dfrac{1}{300} - \dfrac{1}{T_2} \right)$$0.48=19.0960000​(3001​−T2​1​)

$$\dfrac{1}{300} - \dfrac{1}{T_2} = \dfrac{0.48 \times 19.09}{60000}$$3001​−T2​1​=600000.48×19.09​

$$\dfrac{1}{300} - \dfrac{1}{T_2} = \dfrac{9.1632}{60000} = 1.5272 \times 10^{-4}$$3001​−T2​1​=600009.1632​=1.5272×10−4

$$\dfrac{1}{T_2} = \dfrac{1}{300} - 1.5272 \times 10^{-4}$$T2​1​=3001​−1.5272×10−4

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

$$T_2 = \dfrac{1}{3.1806 \times 10^{-3}} \approx 314.4 \text{ K}$$T2​=3.1806×10−31​≈314.4 K

Converting to Celsius:

$$T_2 = 314.4 - 273 = 41.4^\circ\text{C}$$T2​=314.4−273=41.4∘C

The nearest integer is $41$41.

Answer: $41$41