Question:
\(P \rightarrow Q\) is a zero order reaction. If the concentration of \(P\) decreases from \(0.1\ M\) to \(0.05\ M\) in 10 seconds, then the concentration of \(P\) after 15 seconds is
\(P \rightarrow Q\) is a zero order reaction. If the concentration of \(P\) decreases from \(0.1\ M\) to \(0.05\ M\) in 10 seconds, then the concentration of \(P\) after 15 seconds is
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For a zero order reaction,
\[
[A]_t=[A]_0-kt
\]
The concentration decreases linearly with time, and the rate is independent of concentration.
Updated On: Jun 18, 2026
- \(0.05\ M\)
- \(0.025\ M\)
- \(0.02\ M\)
- \(0.01\ M\)
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The Correct Option is B
Solution and Explanation
Step 1: Use the integrated rate equation for a zero order reaction.
For a zero order reaction, \[ [P]_t=[P]_0-kt \] where \[ [P]_0=\text{initial concentration} \] and \[ [P]_t=\text{concentration after time } t \]
Step 2: Calculate the rate constant.
Given, \[ [P]_0=0.10\ M \] and after \(10\ s\), \[ [P]_{10}=0.05\ M \] Using \[ [P]_{10}=[P]_0-k(10) \] \[ 0.05=0.10-10k \] \[ 10k=0.05 \] \[ k=0.005\ M\,s^{-1} \]
Step 3: Calculate the concentration after 15 seconds.
Using \[ [P]_{15}=0.10-(0.005)(15) \] \[ =0.10-0.075 \] \[ =0.025\ M \]
Step 4: Final conclusion.
Therefore, the concentration of \(P\) after \(15\ s\) is \[ \boxed{0.025\ M} \] Hence, the correct option is (2).
For a zero order reaction, \[ [P]_t=[P]_0-kt \] where \[ [P]_0=\text{initial concentration} \] and \[ [P]_t=\text{concentration after time } t \]
Step 2: Calculate the rate constant.
Given, \[ [P]_0=0.10\ M \] and after \(10\ s\), \[ [P]_{10}=0.05\ M \] Using \[ [P]_{10}=[P]_0-k(10) \] \[ 0.05=0.10-10k \] \[ 10k=0.05 \] \[ k=0.005\ M\,s^{-1} \]
Step 3: Calculate the concentration after 15 seconds.
Using \[ [P]_{15}=0.10-(0.005)(15) \] \[ =0.10-0.075 \] \[ =0.025\ M \]
Step 4: Final conclusion.
Therefore, the concentration of \(P\) after \(15\ s\) is \[ \boxed{0.025\ M} \] Hence, the correct option is (2).
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