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
- A 1100
- B 1200
- C 1303
- D 1405
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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