Question

For the reaction \( A + B \to C \), the rate law is found to be \( \text{rate} = k[A]^2[B] \). If the concentration of \( A \) is doubled and \( B \) is halved, by what factor does the rate change?

• A. \( 2 \)
• B. \( \frac{1}{2} \)
• C. \( 4 \)
• D. \( 1 \)

Hint:

When concentrations change, apply exponent in rate law carefully to find new rate.

Answer 1

The Correct Option is A

To ascertain the impact of altered concentrations on the reaction rate, given the rate law \( \text{rate} = k[A]^2[B] \), we must examine the rate's response to a doubling of concentration \( A \) and a halving of concentration \( B \).

Let initial concentrations be \([A]_0\) and \([B]_0\). The initial rate is:

\(\text{rate}_0 = k[A]_0^2[B]_0\)

Following a doubling of \([A]\) to \(2[A]_0\) and a halving of \([B]\) to \(\frac{1}{2}[B]_0\), the new rate, \(\text{rate}_1\), is expressed as:

\(\text{rate}_1 = k(2[A]_0)^2\left(\frac{1}{2}[B]_0\right)\)

The calculation of \(\text{rate}_1\) yields:

\(\text{rate}_1 = k \cdot 4[A]_0^2 \cdot \frac{1}{2}[B]_0 = 2k[A]_0^2[B]_0\)

The factor by which the rate changes is determined by:

\(\frac{\text{rate}_1}{\text{rate}_0} = \frac{2k[A]_0^2[B]_0}{k[A]_0^2[B]_0} = 2\)

Consequently, the reaction rate experiences a twofold increase.

The definitive result is: \( 2 \)