Question

In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L−1 0.20 0.20 0.40 B/ mol L−1 0.30 0.10 0.05 r0/ mol L−1 s−1 5.07 × 10−5 5.07 × 10−5 1.43 × 10−4 What is the order of the reaction with respect to A and B?

Answer 1

The order will be 1.5 with respect to A and 0 with respect to B.

Let the order of the reaction with respect to A be x and with respect to B be y.

• r = k * [A]^x * [B]^y

• 5.07 * 10^(-5) = k * [0.2]^x * [0.3]^y                                  (1)

• 5.07 * 10^(-5) = k * [0.2]^x * [0.1]^y                                   (2)

• 1.43 * 10^(-4) = k * [0.4]^x * [0.05]^y                                 (3)

dividing (1) by (2),

•    1 = [0.3]^y / [0.1]^y

• ⇒ [0.3]^0 / [0.1]^0  = [0.3]^y / [0.1]^y

• ⇒ y = 0

dividing (3) by (2),

• (1.43 * 10^(-4)) / (5.07 * 10^(-5)) = [0.4]^x * [0.05]^y / [0.2]^x * [0.1]^y

since y = 0,

[0.05]^y = [0.3]^y = 1

• ⇒ 2.821 = 2^x

• ⇒ log 2.821 = log 2^x = x log2

• ⇒ x = log2.821 / log2 = 1.496 ≈ 1.5

Answer 2

The order of the reaction with respect to A and B is 1.5

Explanation:

Let x be the order of the reaction with respect to A

Let y be the order of the reaction with respect to B

The order of the reaction is given by the formula:

$\bold{r_{0}=k[A]^{x}[B]^{y}}$

$\bold{5.07 \times 10^{-5}=k[0.20]^{x}[0.30]^{y}\longrightarrow(1)}$

$\bold{5.07 \times 10^{-5}=k[0.20]^{x}[0.10]^{y}\longrightarrow(2)}$

$\bold{1.43 \times 10^{-1}=k[0.40]^{x}[0.05]^{y}\longrightarrow(3)}$

On dividing equation (1) and (2), we get,

$\bold{\frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}}=\frac{k[0.20]^{x}[0.30]^{y}}{k[0.20]^{x}[0.10]^{y}}}$

$\bold{1=\frac{[0.30]^{y}}{[0.10]^{y}}}$

$\bold{\left(\frac{0.30}{0.10}\right)^{0}=\left(\frac{0.30}{0.10}\right)^{y}}$

$\bold{\therefore y = 0}$

On dividing equation (2) and (3), we get,

$\bold{\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{k[0.40]^{x}[0.05]^{y}}{k[0.20]^{x}[0.30]^{y}}}$

On substituting the values of y, we get,

$\bold{\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{[0.40]^{x}}{[0.20]^{x}}}$

Anything power zero is one.

$\bold{2.821=2^{x}}$

On taking log on both sides, we get,

$\bold{\log 2.821=x \log 2}$

$\bold{x=\frac{\log 2.821}{\log 2}}$

$\bold{\therefore x=1.496 \approx 1.5}$