Question:
The unit of rate constant for a first-order reaction is
The unit of rate constant for a first-order reaction is
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Remember that the order of the reaction determines the units of the rate constant. For a zero-order reaction, the units are the same as those of the rate, while for a first-order reaction, the units are inverse time (s$^{-1}$).
Updated On: Jun 3, 2026
- $\text{s}^{-1}$
- $\text{mol L}^{-1}\ \text{s}^{-1}$
- $\text{L mol}^{-1}\ \text{s}^{-1}$
- s
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The rate constant ($k$) for a reaction is a measure of how fast the reaction proceeds. The unit of $k$ depends on the order of the reaction. For a first-order reaction, the rate law can be expressed as: \[\text{Rate} = k[A\] ] where $@@MATH1@@$ is the concentration of reactant A.
Step 2: Meaning
The units of the rate constant for a first-order reaction must match the units of the rate divided by the units of the concentration of the reactant. The rate has units of $\text{mol L}^{-1}\ \text{s}^{-1}$ (concentration per time), and the concentration is in $\text{mol L}^{-1}$. Therefore, to balance the equation: \[\frac{\text{mol L}^{-1}\ \text{s}^{-1}}{\text{mol L}^{-1}} = \text{s}^{-1}\]
Step 3: Analysis
Let's analyze each option: A) $\text{s}^{-1}$: This is the correct unit for a first-order reaction. The rate of the reaction is in $\text{mol L}^{-1}\ \text{s}^{-1}$, and the concentration is in $\text{mol L}^{-1}$. Dividing these units gives $\text{s}^{-1}$. B) $\text{mol L}^{-1}\ \text{s}^{-1}$: This would be the unit for a zero-order reaction where rate = $k@@MATH3@@$, and both have the same concentration units, resulting in $\text{mol L}^{-1}\ \text{s}^{-1}$. C) $\text{L mol}^{-1}\ \text{s}^{-1}$: This is not a valid unit for a first-order reaction. It does not balance the rate law equation correctly. D) s: This is incorrect as it lacks the necessary inverse concentration term to match the units of the rate.
Step 4: Conclusion
The correct unit for the rate constant ($k$) in a first-order reaction is $\text{s}^{-1}$, which matches option A.
Final Answer: (A)
The rate constant ($k$) for a reaction is a measure of how fast the reaction proceeds. The unit of $k$ depends on the order of the reaction. For a first-order reaction, the rate law can be expressed as: \[\text{Rate} = k[A\] ] where $@@MATH1@@$ is the concentration of reactant A.
Step 2: Meaning
The units of the rate constant for a first-order reaction must match the units of the rate divided by the units of the concentration of the reactant. The rate has units of $\text{mol L}^{-1}\ \text{s}^{-1}$ (concentration per time), and the concentration is in $\text{mol L}^{-1}$. Therefore, to balance the equation: \[\frac{\text{mol L}^{-1}\ \text{s}^{-1}}{\text{mol L}^{-1}} = \text{s}^{-1}\]
Step 3: Analysis
Let's analyze each option: A) $\text{s}^{-1}$: This is the correct unit for a first-order reaction. The rate of the reaction is in $\text{mol L}^{-1}\ \text{s}^{-1}$, and the concentration is in $\text{mol L}^{-1}$. Dividing these units gives $\text{s}^{-1}$. B) $\text{mol L}^{-1}\ \text{s}^{-1}$: This would be the unit for a zero-order reaction where rate = $k@@MATH3@@$, and both have the same concentration units, resulting in $\text{mol L}^{-1}\ \text{s}^{-1}$. C) $\text{L mol}^{-1}\ \text{s}^{-1}$: This is not a valid unit for a first-order reaction. It does not balance the rate law equation correctly. D) s: This is incorrect as it lacks the necessary inverse concentration term to match the units of the rate.
Step 4: Conclusion
The correct unit for the rate constant ($k$) in a first-order reaction is $\text{s}^{-1}$, which matches option A.
Final Answer: (A)
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