AIIMS 2005 Chemistry Order of Reaction MCQ Question

To determine the order of the reaction with respect to reactant A in the given problem, we use the relationship between the rate of the reaction and the concentration of reactants, which can be expressed with the rate equation:

$$\text{Rate} = k[A]^n$$

where:

• $k$ is the rate constant,
• $[A]$ is the concentration of reactant A,
• $n$ is the order of the reaction with respect to A.

Given Data:

1. When $[A] = 2.2 \, \text{mM}$, the rate $\text{Rate}_1 = 2.4 \, \text{mM s}^{-1}$.
2. When $[A] = 1.1 \, \text{mM}$ (which is half of 2.2 mM), the rate $\text{Rate}_2 = 0.6 \, \text{mM s}^{-1}$.

Step 1: Set Up the Rate Equations

From the first condition:

$$\text{Rate}_1 = k [A]_1^n \implies 2.4 = k (2.2)^n \quad (1)$$

From the second condition:

$$\text{Rate}_2 = k [A]_2^n \implies 0.6 = k (1.1)^n \quad (2)$$

Step 2: Divide the Two Equations

To eliminate $k$, we can divide equation (1) by equation (2):

$$\frac{2.4}{0.6} = \frac{k (2.2)^n}{k (1.1)^n}$$

This simplifies to:

$$4 = \left(\frac{2.2}{1.1}\right)^n$$

Since $\frac{2.2}{1.1} = 2$, we can rewrite the equation as:

$$4 = 2^n$$

Step 3: Solve for $n$

To find $n$, we recognize that:

$$4 = 2^2 \implies n = 2$$

Conclusion

The order of reaction with respect to A is $n = 2$.

Why This Answer is Correct

The calculations show that when the concentration of A is halved, the rate of reaction decreases by a factor of 4, which aligns perfectly with a second-order reaction. In a second-order reaction, if the concentration of reactant is halved, the rate decreases by the square of the concentration change, confirming that the order of reaction is indeed 2.

Why Other Options are Incorrect

• Option A (1.5): If the order were 1.5, the rate would not decrease by a factor of 4 when the concentration is halved.
• Option C (2.5): A non-integer order like 2.5 would yield inconsistent results with the observed rate changes.
• Option D (3.0): If the order were 3, we would expect the rate to decrease by a factor of 8 when the concentration is halved, which is not the case here.

Thus, the correct answer is B) 2.0.