Question:
For a second-order reaction, the unit of rate constant is:
For a second-order reaction, the unit of rate constant is:
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Common units:
\[
0^{th}: molL^{-1}s^{-1}
\]
\[
1^{st}: s^{-1}
\]
\[
2^{nd}: Lmol^{-1}s^{-1}
\]
Updated On: Jun 8, 2026
- \(s^{-1}\)
- \(mol\,L^{-1}s^{-1}\)
- \(L\,mol^{-1}s^{-1}\)
- \(L^2mol^{-2}s^{-1}\)
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The Correct Option is C
Solution and Explanation
Concept:
For an \(n^{th}\) order reaction:
\[
[k]
=
(\text{concentration})^{1-n}
(\text{time})^{-1}
\]
Step 1: Write rate law.
For second-order reaction: \[ Rate=k[A]^2 \]
Step 2: Substitute units.
Rate: \[ molL^{-1}s^{-1} \] Concentration: \[ molL^{-1} \] Therefore: \[ k= \frac{molL^{-1}s^{-1}} {(molL^{-1})^2} \] \[ = Lmol^{-1}s^{-1} \]
Step 3: Conclusion.
\[ \boxed{Lmol^{-1}s^{-1}} \]
Step 1: Write rate law.
For second-order reaction: \[ Rate=k[A]^2 \]
Step 2: Substitute units.
Rate: \[ molL^{-1}s^{-1} \] Concentration: \[ molL^{-1} \] Therefore: \[ k= \frac{molL^{-1}s^{-1}} {(molL^{-1})^2} \] \[ = Lmol^{-1}s^{-1} \]
Step 3: Conclusion.
\[ \boxed{Lmol^{-1}s^{-1}} \]
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