Question

For the reaction: \[ 2A + B \rightarrow 3C \] The rate law is given as: \[ \text{Rate} = k[A]^2[B] \] If the concentration of A is doubled and the concentration of B is halved, how does the rate of the reaction change?

• A. It remains the same.
• B. It increases by a factor of 4.
• C. It doubles.
• D. It decreases by a factor of 2.

Hint:

In chemical kinetics, the rate law describes how the rate of a reaction depends on the concentrations of reactants. Pay attention to the powers of the concentration terms in the rate law when altering reactant concentrations.

Answer 1

The Correct Option is C

The rate law is expressed as:

\[ \text{Rate} = k[A]^2[B] \]

Step 1: Initial Rate Determination

With initial concentrations denoted as \( [A]_0 \) and \( [B]_0 \), the initial rate is:

\[ \text{Rate}_0 = k[A]_0^2[B]_0 \]

Step 2: Updated Concentrations

The concentration of \( A \) is doubled:

\[ [A]_{\text{new}} = 2[A]_0 \]

The concentration of \( B \) is halved:

\[ [B]_{\text{new}} = \frac{1}{2}[B]_0 \]

Step 3: New Rate Calculation

Substituting the updated concentrations into the rate law yields:

\[ \text{Rate}_{\text{new}} = k(2[A]_0)^2\left(\frac{1}{2}[B]_0\right) \]

\[ \text{Rate}_{\text{new}} = k \cdot 4[A]_0^2 \cdot \frac{1}{2}[B]_0 \]

Simplification results in:

\[ \text{Rate}_{\text{new}} = 2k[A]_0^2[B]_0 \]

Step 4: Rate Comparison

The initial rate was established as:

\[ \text{Rate}_0 = k[A]_0^2[B]_0 \]

The newly calculated rate is:

\[ \text{Rate}_{\text{new}} = 2 \cdot \text{Rate}_0 \]

✅ Final Answer:

The reaction rate doubles. The correct response is \( \boxed{\text{It doubles.}} \).