Read the Passage carefully and answer the Questions.
Kinetic studies not only help us to determine the rate of a chemical reaction but also describe the conditions by which the reaction rates can be altered. The factors such as concentration, temperature, pressure and catalyst affect the rate of a reaction. The speed of a reaction or the rate of a reaction can be defined as the change in concentration of a reactant or product in unit time.
Rate = $k [A]^x [B]^y$
Here k is the rate constant
$x + y$ gives the overall order of a reaction, whereas x and y represent the order with respect to the reactants A and B, respectively.
Table: Initial rate of formation of $NO_2$
|
S.No. |
Initial $[NO]/mol\, L^{-1}$ |
Initial $[O_2]/mol\, L^{-1}$ |
Initial rate of formation of $NO_2/mol\, L^{-1}\, s^{-1}$ |
|
1 |
0.30 |
0.30 |
0.096 |
|
2 |
0.60 |
0.30 |
0.384 |
|
3 |
0.30 |
0.60 |
0.192 |
|
4 |
0.60 |
0.60 |
0.768 |
Based on the above observations, the rate of formation of $NO_2$ can be represented as
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Rate of reaction = $k[NO]^2 [O_2]$
The rate of formation of $\text{NO}_2$ can be represented by the general rate law:
$\text{Rate} = k[\text{NO}]^x[\text{O}_2]^y$
where $x$ and $y$ are the reaction orders with respect to $\text{NO}$ and $\text{O}_2$, respectively.
We determine $x$ and $y$ using the method of initial rates.
1. Determine the Order with Respect to $\text{NO}$ ($x$)
We compare experiments where the concentration of $\text{O}_2$ is held constant (Experiments 1 and 2).
$\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[\text{NO}]_2^x[\text{O}_2]_2^y}{k[\text{NO}]_1^x[\text{O}_2]_1^y}$
$\frac{0.384}{0.096} = \left(\frac{0.60}{0.30}\right)^x \left(\frac{0.30}{0.30}\right)^y$
$4 = (2)^x$
$\mathbf{x = 2}$
The reaction is second order with respect to $\text{NO}$.
2. Determine the Order with Respect to $\text{O}_2$ ($y$)
We compare experiments where the concentration of $\text{NO}$ is held constant (Experiments 1 and 3).
$\frac{\text{Rate}_3}{\text{Rate}_1} = \frac{k[\text{NO}]_3^x[\text{O}_2]_3^y}{k[\text{NO}]_1^x[\text{O}_2]_1^y}$
$\frac{0.192}{0.096} = \left(\frac{0.30}{0.30}\right)^x \left(\frac{0.60}{0.30}\right)^y$
$2 = (2)^y$
$\mathbf{y = 1}$
The reaction is first order with respect to $\text{O}_2$.
3. Write the Rate Law
Substituting the orders ($x=2$ and $y=1$) into the general rate law:
$\text{Rate of reaction} = k[\text{NO}]^2[\text{O}_2]$
The correct option is: $\text{Rate of reaction} = k[\text{NO}]^2[\text{O}_2]$