Question

The reaction A + B → C is elementary. If the concentration of A is doubled, the rate will:

• A. Remain the same
• B. Double
• C. Quadruple
• D. Decrease by half

Hint:

For elementary reactions, the exponents in the rate law match the stoichiometric coefficients. If a reactant's concentration is doubled, the rate changes accordingly.

Answer 1

The Correct Option is B

For an elementary reaction, the rate law can be written directly from the stoichiometry of the reaction.
Given the reaction: \[ \text{A} + \text{B} \rightarrow \text{C} \]
This implies that the reaction is first order with respect to A and first order with respect to B. Therefore, the rate law is:
\[ \text{Rate} = k[A][B] \]
If the concentration of A is doubled (i.e., $[A] \rightarrow 2[A]$), while $[B]$ remains constant, then:
\[ \text{New Rate} = k(2[A])[B] = 2k[A][B] = 2 \times \text{Original Rate} \]
Thus, the reaction rate doubles.
Now consider the other options:
- (1) Remain the same: Incorrect, since the rate depends on $[A]$.
- (3) Quadruple: Would happen if both $[A]$ and $[B]$ were doubled.
- (4) Decrease by half: Irrelevant for increasing $[A]$ — not correct here.
Hence, for an elementary reaction where only $[A]$ is doubled, the rate also doubles.