Q127 / 200 · MHT CET - Medical 2026 (22 April Shift 2) · Chemistry
The rate law for the reaction, $\text{A} + \text{B} \rightarrow \text{P}$ is rate $= k[\text{A}][\text{B}]$. The rate constant of the reaction is $4.5\ \text{M}^{-1}\text{s}^{-1}$ at $25^\circ$C. Calculate the rate of the reaction when $[\text{A}] = 0.1$ M and $[\text{B}] = 0.2$ M.
- A$9.25 \times 10^{-2}\ \text{M s}^{-1}$
- Correct answerB$9.0 \times 10^{-2}\ \text{M s}^{-1}$
- C$9.50 \times 10^{-2}\ \text{M s}^{-1}$
- D$9.75 \times 10^{-2}\ \text{M s}^{-1}$
Solution
Given the rate law for the reaction $\text{A} + \text{B} \rightarrow \text{P}$ is:
$\text{Rate} = k[A][B]$
Substituting the given values:
$k = 4.5 \text{ M}^{-1}\text{s}^{-1}$
$[A] = 0.1 \text{ M}$
$[B] = 0.2 \text{ M}$
$\text{Rate} = 4.5 \times 0.1 \times 0.2$
$\text{Rate} = 4.5 \times 0.02$
$\text{Rate} = 0.090 \text{ M s}^{-1}$
Answer: $9.0 \times 10^{-2}\ \text{M s}^{-1}$
$\text{Rate} = k[A][B]$
Substituting the given values:
$k = 4.5 \text{ M}^{-1}\text{s}^{-1}$
$[A] = 0.1 \text{ M}$
$[B] = 0.2 \text{ M}$
$\text{Rate} = 4.5 \times 0.1 \times 0.2$
$\text{Rate} = 4.5 \times 0.02$
$\text{Rate} = 0.090 \text{ M s}^{-1}$
Answer: $9.0 \times 10^{-2}\ \text{M s}^{-1}$
Hence, the correct answer is option: 'B'.