Question

The rate law equation for A \(\rightarrow\) Product, is rate = k[A]\(^x\). What is the effect of increase in concentration of 'A' on rate of reaction, if x<0?

Answer 1

Step 1: Recall the rate law.
The general form of the rate law is: \[ \text{Rate} = k[A]^x \] where:

• \(k\) = rate constant
• \([A]\) = concentration of reactant A
• \(x\) = order of reaction with respect to A

Step 2: Case of negative order.
If \(x < 0\), then the concentration of A appears in the denominator: \[ \text{Rate} = k \cdot \frac{1}{[A]^{|x|}} \] where \(|x|\) represents the absolute value of the negative exponent.

Step 3: Effect of increasing [A].
Since \([A]\) is in the denominator, as the concentration of A increases, the overall rate of the reaction decreases. 
Conversely, if \([A]\) decreases, the rate of the reaction increases.

Step 4: Conclusion.
For a reaction with negative order in A, the reactant A behaves like an inhibitor. 
Therefore, increasing \([A]\) reduces the reaction rate.

Final Answer:

If \(x < 0\), an increase in concentration of A will decrease the rate of the reaction.