Question

For a given reaction, $ X(g) + Y(g) \to Z(g) $, the order of reaction with respect to $ X $ and $ Y $ are $ m $ and $ n $ respectively. If the concentration of $ X $ is tripled and that of $ Y $ is decreased to one third, what is the ratio between the new rate to the original rate of the reaction?

• A. \( 3^{(m-n)} \)
• B. \( 3^{(m)} \)
• C. \( (m + n) \)
• D. \( \frac{1}{3^{(m+n)}} \)

Hint:

In problems involving changes in concentration, use the rate law to express how the rate is affected by the changes in concentration. Pay attention to how the changes in concentration are reflected in the exponents.

Answer 1

The Correct Option is A

The rate law for the reaction is given by: \[ \text{Rate} = k [X]^m [Y]^n \] Where \( k \) is the rate constant, \( m \) is the order with respect to \( X \), and \( n \) is the order with respect to \( Y \). Now, the initial rate of the reaction is: \[ \text{Initial rate} = k [X]^m [Y]^n \] Next, the concentrations of \( X \) and \( Y \) are changed as follows: - The concentration of \( X \) is tripled, so the new concentration is \( 3[X] \). - The concentration of \( Y \) is reduced to one third, so the new concentration is \( \frac{1}{3}[Y] \). The new rate will be: \[ \text{New rate} = k (3[X])^m \left(\frac{1}{3}[Y]\right)^n = k 3^m [X]^m \cdot \frac{1}{3^n} [Y]^n \] Thus, the ratio of the new rate to the initial rate is: \[ \frac{\text{New rate}}{\text{Initial rate}} = \frac{k 3^m [X]^m \cdot \frac{1}{3^n} [Y]^n}{k [X]^m [Y]^n} = 3^{m-n} \] Therefore, the ratio between the new rate and the original rate is \( 3^{(m-n)} \). Thus, the correct answer is \( (A) \).