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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2026 · 22 January · Morning Shift · Chemistry · Chemical Kinetics

The temperature at which the rate constants of the given below two...

The temperature at which the rate constants of the given below two gaseous reactions become equal is ____\_\_\_\_ K. (Nearest integer)
X⟶Yk1=106e−30000 T\mathrm{X} \longrightarrow \mathrm{Y} \quad \mathrm{k}_{1}=10^{6} e^{\frac{-30000}{\mathrm{~T}}}
P⟶Qk2=104e−24000 T\mathrm{P} \longrightarrow \mathrm{Q} \quad \mathrm{k}_{2}=10^{4} e^{\frac{-24000}{\mathrm{~T}}}
Given: ln⁡10=2.303\ln 10=2.303

Correct answer

0

Step-by-step solution

To find the temperature at which k1=k2k_1 = k_2,
set the rate constant equations equal: 106e−30000/T=104e−24000/T10^6 e^{-30000/T} = 10^4 e^{-24000/T}.
Dividing both sides by 10410^4: 100=e−24000/T+30000/T=e6000/T100 = e^{-24000/T + 30000/T} = e^{6000/T}.
Taking natural logarithm: ln⁡(100)=6000T\ln(100) = \frac{6000}{T}.
Since ln⁡(100)=2ln⁡(10)=2×2.303=4.606\ln(100) = 2\ln(10) = 2 \times 2.303 = 4.606: 4.606=6000T4.606 = \frac{6000}{T}.
Solving for T: T=60004.606=1302.5T = \frac{6000}{4.606} = 1302.5 K.
Rounding to the nearest integer gives T = 1303 K.
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