\(P \rightarrow Q\) 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

Question

Following are the four molecules ``P``, ``Q``, ``R`` and ``S``. Which one among the four molecules will react with H-Br(aq) at the fastest rate? Molecules:

\[ P: \text{Cyclic compound with two O groups attached to the ring.} \] \[ Q: \text{Cyclic compound with one O group and one CH\(_3\) group attached to the ring.} \] \[ R: \text{Cyclic compound with one O group attached to the ring and one CH\(_3\) group attached to the ring.} \] \[ S: \text{Cyclic compound with one CH\(_3\) group attached to the ring.} \]

Answer 1

Step 1: Extract k from given zero-order data.
[P] drops from 0.10 to 0.05 M in 10 s → k = (0.10–0.05)/10 = 0.005 M/s.

Step 2: Project concentration forward.
At t=15 s: [P] = 0.10 – 0.005×15 = 0.025 M.

Step 3: Final Answer:
0.025 M (option 2).

\(P \rightarrow Q\) 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

Show Hint

The Correct Option is B

Solution and Explanation

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