Question:
For the two gaseous reactions, following data:
\(A \rightarrow B; k_1 = 10^{10}e^{-20,000/T}\)
\(C \rightarrow D; k_2 = 10^{12}e^{-24,606/T}\)
the temperature at which \(k_1\) becomes equal to \(k_2\) is
For the two gaseous reactions, following data:
\(A \rightarrow B; k_1 = 10^{10}e^{-20,000/T}\)
\(C \rightarrow D; k_2 = 10^{12}e^{-24,606/T}\)
the temperature at which \(k_1\) becomes equal to \(k_2\) is
\(A \rightarrow B; k_1 = 10^{10}e^{-20,000/T}\)
\(C \rightarrow D; k_2 = 10^{12}e^{-24,606/T}\)
the temperature at which \(k_1\) becomes equal to \(k_2\) is
Show Hint
Use \(\ln 10 = 2.303\) and \(\ln 100 = 4.606\).
Updated On: Apr 20, 2026
- 400 K
- 1000 K
- 800 K
- 1500 K
- 500 K
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Set \(k_1 = k_2\) and solve for T.
Step 2: Detailed Explanation:
\(10^{10}e^{-20,000/T} = 10^{12}e^{-24,606/T}\)
\(e^{-20,000/T} / e^{-24,606/T} = 10^{12}/10^{10}\)
\(e^{(24,606-20,000)/T} = 10^2\)
\(e^{4,606/T} = 100\)
Taking natural log: \(4,606/T = \ln 100 = 2 \times 2.303 = 4.606\)
\(T = 4,606 / 4.606 = 1000\) K.
Step 3: Final Answer:
1000 K
Set \(k_1 = k_2\) and solve for T.
Step 2: Detailed Explanation:
\(10^{10}e^{-20,000/T} = 10^{12}e^{-24,606/T}\)
\(e^{-20,000/T} / e^{-24,606/T} = 10^{12}/10^{10}\)
\(e^{(24,606-20,000)/T} = 10^2\)
\(e^{4,606/T} = 100\)
Taking natural log: \(4,606/T = \ln 100 = 2 \times 2.303 = 4.606\)
\(T = 4,606 / 4.606 = 1000\) K.
Step 3: Final Answer:
1000 K
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