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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

The temperature at which the rate constants of the given below two gaseous reactions become equal is _______ K. (Nearest integer).
\(X \longrightarrow Y \quad k _1=10^6 e ^{\frac{-30000}{T}}\)
\(P \longrightarrow Q \quad k _2=10^4 e ^{\frac{-24000}{T}}\)
Given : ln 10 = 2.303

  1. A 1100
  2. B 1200
  3. C 1303
  4. D 1405
Verified Solution

Answer & Solution

Correct Answer

(C) 1303

Step-by-step Solution

Detailed explanation

\(10^4 e ^{\frac{-24000}{T}}=10^6 e ^{\frac{-30000}{T}}\) \(e ^{\frac{6000}{T}}=100\) \(\frac{6000}{T}=2 \ln 10\) \(T =\frac{6000}{2 \times 2.303}\) \(T =1302.64 K\) \(T \approx 1303 K\)
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