(a) Determination of Order of Reaction

Step 1 (Formula & Given Data): [0.5 Mark]

The rate law for a reaction with respect to reactant 'X' is given by:

$$Rate = k[X]^n$$

Where $n$ is the order of the reaction. Let the initial rate be $R_1 = k[X]^n$ and the new rate be $R_2 = k[3X]^n$. Given that $R_2 = 27 R_1$.

Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]

Substituting the values into the ratio:

$$\frac{R_2}{R_1} = \frac{k[3X]^n}{k[X]^n}$$ $$27 = \frac{3^n \cdot [X]^n}{[X]^n}$$ $$27 = 3^n$$ $$3^3 = 3^n$$

Step 3 (Final Concluded Answer): [0.0 Mark]

Comparing the exponents, we get $n = 3$. The order of the reaction is 3 (Third order).

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(b) Pseudo First-Order Reaction

Step 1 (Condition & Definition): [0.5 Mark]

A bimolecular reaction behaves as a first-order reaction when one of the reactants is present in large excess. Because its concentration remains practically constant throughout the reaction, it does not appear in the rate expression.

Step 2 (Example): [0.5 Mark]

The acid-catalyzed hydrolysis of ethyl acetate is a classic example:

$$CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH$$

The rate law is: $Rate = k[CH_3COOC_2H_5]^1[H_2O]^0$, which simplifies to $Rate = k'[CH_3COOC_2H_5]$.

💡 EL10 Concept Insight:

Think of it like adding a drop of ink to a giant swimming pool; the ink changes the color, but the pool's total volume stays basically the same. In chemistry, when you have so much of one ingredient that its amount barely changes, the reaction speed only depends on the other ingredient, making it look like a simple first-order reaction!