Step 1 (Formula / Governing Law & Given Data): [1 Mark]
The rate law for the reaction $\text{A} + \text{B} \rightarrow \text{Products}$ is expressed as:
$$\text{Rate} = k[\text{A}]^x[\text{B}]^y$$Where $x$ and $y$ are the orders of reaction with respect to A and B, respectively. From the given data table:
- Exp 1: $0.05 = k(0.1)^x(0.1)^y$
- Exp 2: $0.10 = k(0.2)^x(0.1)^y$
- Exp 3: $0.05 = k(0.1)^x(0.2)^y$
Step 2 (Step-by-Step Mathematical Working): [1.5 Marks]
To find order with respect to A ($x$): Divide Exp 2 by Exp 1:
$$\frac{0.10}{0.05} = \frac{k(0.2)^x(0.1)^y}{k(0.1)^x(0.1)^y}$$ $$2 = \left(\frac{0.2}{0.1}\right)^x \implies 2 = 2^x \implies x = 1$$To find order with respect to B ($y$): Divide Exp 3 by Exp 1:
$$\frac{0.05}{0.05} = \frac{k(0.1)^x(0.2)^y}{k(0.1)^x(0.1)^y}$$ $$1 = \left(\frac{0.2}{0.1}\right)^y \implies 1 = 2^y$$Since any number raised to the power of 0 is 1, $y = 0$.
Step 3 (Final Concluded Answer): [0.5 Marks]
- Order with respect to A: 1
- Order with respect to B: 0
- Overall order of the reaction ($x + y$): 1 + 0 = 1
💡 EL10 Concept Insight: Think of it like a recipe: if doubling the amount of ingredient A doubles the speed of cooking, it's a 1st-order reaction; but if doubling ingredient B doesn't change the speed at all, it means B doesn't affect the reaction rate, making it a 0th-order reaction.