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 the problem, we need to understand how the rate of the reaction changes when the concentration of reactant A is altered in a rate law expressed as \( \text{rate} = k[A]^2[B] \).

Here, \( k \) is the rate constant, \([A]\) is the concentration of reactant A, and \([B]\) is the concentration of reactant B.

The rate law indicates the reaction is second-order with respect to \([A]\) and first-order with respect to \([B]\).

Let's consider the initial rate \(\text{rate}_0\) when \([A] = a\) and \([B] = b\):

\(\text{rate}_0 = k[a]^2[b]\) 

When the concentration of \(A\) is tripled, \([A]\) becomes \(3a\), then the new rate \(\text{rate}_\text{new}\) is:

\(\text{rate}_\text{new} = k(3a)^2[b]\)

Simplifying gives:

\(\text{rate}_\text{new} = k \cdot 9a^2 \cdot b = 9 \cdot k[a]^2[b]\)

Therefore, \(\text{rate}_\text{new} = 9 \cdot \text{rate}_0\).

This means the initial rate increases by a factor of nine when the concentration of A is tripled, confirming the increase is by a factor of nine.

Answer 2

Rate R = k [A]²[B] 
if [A] is tripled and [B] stays the same. 
Then R' = k [3A]² [B]  
R' = 9k [A]² [B]  
R' = 9R
Hence, the initial rate is increased by nine times.
Therefore, The correct option is (B): increase by a factor of nine.