Question

For a certain reaction, the rate=\(k[A]^2[B]\) when the initial concentration of A is tripled keeping concentration of B constant, the initial rate would

• A. increase by a factor of six.
• B. increase by a factor of nine.
• C. increase by a factor of three.
• D. decrease by a factor of nine.

Answer 1

The Correct Option is B

To solve this problem, we need to understand the effect of changing the concentration of reactants on the rate of a reaction that follows a specific rate law.

The given rate law for the reaction is:

\(rate = k[A]^2[B]\)

Initially, let the concentration of A be \([A]_0\) and the concentration of B be \([B]_0\). The initial rate can then be written as:

\(rate_{initial} = k[A]_0^2[B]_0\)

When the concentration of A is tripled, the new concentration of A becomes \(3[A]_0\), while the concentration of B remains constant at \([B]_0\).

The new rate of the reaction can be calculated as follows:

\(rate_{new} = k(3[A]_0)^2[B]_0\)

Expanding this expression gives:

\(rate_{new} = k \cdot 9[A]_0^2[B]_0\)

We can compare this new rate with the initial rate:

\(\frac{rate_{new}}{rate_{initial}} = \frac{k \cdot 9[A]_0^2[B]_0}{k[A]_0^2[B]_0}\)

The \(k\), \([A]_0^2\), and \([B]_0\) terms cancel out, simplifying the expression to:

\(\frac{rate_{new}}{rate_{initial}} = 9\)

This shows that the initial rate increases by a factor of nine when the concentration of A is tripled and the concentration of B is kept constant.

Therefore, the correct answer is: increase by a factor of nine.