Question

The experimental data for the reaction given below are as follows. Write the probable rate expressions.
Reaction: $2A+{{B}_{2}}\to 2AB$

$[A]$ $mol\text{ }litr{{e}^{-1}}$	$[{{B}_{2}}]$ $mol\text{ }litr{{e}^{-1}}$	$Rate\times {{10}^{4}}$ $mol\text{ }litr{{e}^{-1}}{{s}^{-1}}$
0.50	0.50	1.6
0.50	1.00	3.2
1.00	1.00	3.2

Answer 1

Hint: Consider the equation for the rate of a reaction and how it depends on the concentration of the reactants. Find the power to which we will raise the concentrations of the reactants to find the probable rate expression.

Compete step by step solution:
We know that the general expression for the rate of a reaction includes the rate constant and the product of the concentrations of the reactants raised to a certain power which is known as the partial order of the reaction. It is shown as follows:
\[r=k{{[A]}^{m}}{{[B]}^{n}}\]
Here, $r$ is the rate, $k$ is the rate constant, $A$ and $B$ are the reactants, and $m$ and $n$ are the partial orders. We will find the values of the partial orders for this reaction from the given information.
The equations for these reactions will be as follows:
\[\begin{align}
& i)1.6\times {{10}^{4}}=k{{[0.5]}^{m}}{{[0.5]}^{n}} \\
& ii)3.2\times {{10}^{4}}=k{{[0.5]}^{m}}{{[1.0]}^{n}} \\
& iii)3.2\times {{10}^{4}}=k{{[1.0]}^{m}}{{[1.0]}^{n}} \\
\end{align}\]
We will divide equation ii) by equations i) and iii) to obtain the values of n and m respectively.
- For n
\[\begin{align}
&\implies \dfrac{3.2\times {{10}^{4}}}{1.6\times {{10}^{4}}}=\dfrac{k{{[0.5]}^{m}}{{[1.0]}^{n}}}{k{{[0.5]}^{m}}{{[0.5]}^{n}}} \\
&\implies 2={{\left( \dfrac{1.0}{0.5} \right)}^{n}} \\
& n=1 \\
\end{align}\]
- For m
\[\begin{align}
&\implies \dfrac{3.2\times {{10}^{4}}}{3.2\times {{10}^{4}}}=\dfrac{k{{[0.5]}^{m}}{{[1.0]}^{n}}}{k{{[1.0]}^{m}}{{[1.0]}^{n}}} \\
&\implies 1={{\left( \dfrac{0.5}{1.0} \right)}^{m}} \\
& m=0 \\
\end{align}\]
So, the values of m and n are 0 and 1 respectively. Now, putting these in the general rate equation, we get the probable rate equation as:
\[\begin{align}
& r=k{{[A]}^{0}}{{[{{B}_{2}}]}^{1}} \\
& r=k[B] \\
\end{align}\]
We can infer that this is a first order reaction whose rate depends only on the concentration of the reactant ${{B}_{2}}$.

Note: Remember that if no such information about the rate of the reaction in certain cases is given, we have to assume that these partial orders of the reaction are the stoichiometric coefficients of the respective reactants. This may not always be true as we can see here that the stoichiometric coefficient of $A$ is 2 but its partial order is 0.