Question

For the reaction : 2CO(g) + O2(g) → 2CO2(g)

Following data is given :

The overall order of reaction is -

• A. 0
• B. 1
• C. 2
• D. -1

Answer 1

Answer: (A)

0

Answer 2

The reaction is given as: $2CO(g) + O_2(g) \rightarrow 2CO_2(g)$

The rate law for the reaction can be expressed as: $\text{Rate} = k[CO]^x[O_2]^y$

where 'x' is the order of the reaction with respect to CO, 'y' is the order of the reaction with respect to $O_2$, and 'k' is the rate constant. The overall order of the reaction is $x+y$.

We use the provided experimental data to determine the values of 'x' and 'y'.

1. Determine the order with respect to CO (x):

Compare experiments 1 and 2, where the concentration of $O_2$ is kept constant.

From Experiment 1: $4 \times 10^{-5} = k[0.02]^x[0.02]^y$ (Equation 1)

From Experiment 2: $8 \times 10^{-5} = k[0.04]^x[0.02]^y$ (Equation 2)

Divide Equation 2 by Equation 1:

$\frac{8 \times 10^{-5}}{4 \times 10^{-5}} = \frac{k[0.04]^x[0.02]^y}{k[0.02]^x[0.02]^y}$

$2 = \left(\frac{0.04}{0.02}\right)^x$

$2 = (2)^x$

Therefore, $x = 1$.

The reaction is first order with respect to CO.

2. Determine the order with respect to $O_2$ (y):

Compare experiments 1 and 3, where the concentration of CO is kept constant.

From Experiment 1: $4 \times 10^{-5} = k[0.02]^x[0.02]^y$ (Equation 1)

From Experiment 3: $2 \times 10^{-5} = k[0.02]^x[0.04]^y$ (Equation 3)

Divide Equation 3 by Equation 1:

$\frac{2 \times 10^{-5}}{4 \times 10^{-5}} = \frac{k[0.02]^x[0.04]^y}{k[0.02]^x[0.02]^y}$

$0.5 = \left(\frac{0.04}{0.02}\right)^y$

$\frac{1}{2} = (2)^y$

$2^{-1} = (2)^y$

Therefore, $y = -1$.

The reaction is minus first order with respect to $O_2$.

3. Calculate the overall order of the reaction:

Overall order = $x + y$

Overall order = $1 + (-1)$

Overall order = $0$

The overall order of the reaction is 0.