Chemistry Reaction Kinetics MDCAT 2015
PMDC Verified Question 71 of 80
When the change in concentration is \( 6 \times 10^{-4} \text{ mol dm}^{-3} \) and time for that change is 10 seconds, the rate of reaction will be?
A
\( 6 \times 10^{-3} \text{ mol dm}^{-3} \text{sec}^{-1} \)
B
\( 6 \times 10^{-4} \text{ mol dm}^{-3} \text{sec}^{-1} \)
C
\( 6 \times 10^{-2} \text{ mol dm}^{-3} \text{sec}^{-1} \)
D
\( 6 \times 10^{-5} \text{ mol dm}^{-3} \text{sec}^{-1} \)
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Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: \( 6 \times 10^{-5} \text{ mol dm}^{-3} \text{sec}^{-1} \)
Concept:

The rate of a chemical reaction is defined as the change in concentration of a reactant or product per unit of time.

Formula:

$$ \text{Rate} = \frac{\Delta C}{\Delta t} $$

Solution:

  • Given change in concentration, \( \Delta C = 6 \times 10^{-4} \text{ mol dm}^{-3} \).
  • Given time interval, \( \Delta t = 10 \text{ seconds} \).
  • Substitute the values: $$ \text{Rate} = \frac{6 \times 10^{-4}}{10} $$
  • $$ \text{Rate} = 6 \times 10^{-5} \text{ mol dm}^{-3} \text{sec}^{-1} $$


Why other options are incorrect:

Option A results from incorrectly multiplying by 10 instead of dividing. The others are simple arithmetic errors.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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