Question

For the reaction \(2\mathrm{A} + \mathrm{B} \to \mathrm{C}\), the values of initial rate at diffrent reactant concentrations are given in the table below. The rate law for the reaction is :

[A] (mol L-1)	[B] (mol L-1)	Initial Rate (mol L-1 s-1)
0.05	0.05	0.045
0.10	0.05	0.090
0.20	0.10	0.72

• A. \(\mathrm{Rate} = k[\mathrm{A}][\mathrm{B}]^2\)
• B. \(\mathrm{Rate} = k[\mathrm{A}]^2[\mathrm{B}]^2\)
• C. \(\mathrm{Rate} = k[\mathrm{A}]^2[\mathrm{B}]\)
• D. \(\mathrm{Rate} = k[\mathrm{A}][\mathrm{B}]\)

Hint:

Compare rows 1 and 2 (only [A] changes) to get the order in A, then use row 3 (both change) to get the order in B.

Answer 1

The Correct Option is A

Let Rate \(= k[\mathrm{A}]^x[\mathrm{B}]^y\).
Rows 1 and 2: [B] is constant (0.05), [A] doubles (0.05 to 0.10) and the rate doubles (0.045 to 0.090). So \(2^x = 2\), \(x = 1\).
Rows 2 and 3: [A] doubles (0.10 to 0.20), [B] doubles (0.05 to 0.10), rate becomes \(0.72/0.090 = 8\) times. So \(2^x \cdot 2^y = 8\) with \(x = 1\): \(2^y = 4\), \(y = 2\).
Rate law: \(\mathrm{Rate} = k[\mathrm{A}][\mathrm{B}]^2\).
Answer: (1) \(\mathrm{Rate} = k[\mathrm{A}][\mathrm{B}]^2\)