Question

Select the rate law that corresponds to the data shown for the following reaction:
$A + B \to C$ .

EXPERIMENT NO.	(1)	(2)	INITIAL RATE
1	0.012	0.035	0.10
2	0.024	0.070	0.80
3	0.024	0.035	0.10
4	0.012	0.070	0.80

Answer 1

The rate law is given by the formula :
$R = k{\left[ A \right]^x}{\left[ B \right]^y}$
First we’ll put the values of any two experiments in such a way so that any one concentration would get cancelled. Then, in a similar way we’ll substitute again the values of any two experiments. In this way, x and y can be calculated.

Complete step by step answer: The rate law or rate equation for a chemical reaction is an equation that links the initial or forward reaction rate with the concentrations or pressures of the reactants. For many reactions, the initial rate $\left( R \right)$ is given by a power law such as
$R = k{\left[ A \right]^x}{\left[ B \right]^y}$
where$\left[ A \right]$ and $\left[ B \right]$ express the concentration of the solution A and B, the exponents x and y are the partial orders of the reaction for A and B, k is known as reaction rate constant or rate coefficient of the reaction.
Hence, the rate law is given by the formula:
$R = k{\left[ A \right]^x}{\left[ B \right]^y}$
Column 1 represents experiment numbers, column 2 and 3 represents concentration of A and B at different experiments and column 4 represents the initial rate of the reaction. So, expt.1 and 3 can be written as:
$0.10 = k{\left[ {0.012} \right]^x}{\left[ {0.035} \right]^y}$ and$0.10 = k{\left[ {0.024} \right]^x}{\left[ {0.035} \right]^y}$
Now, we’ll divide both of these rate laws i.e.,
$\dfrac{{0.10}}{{0.10}} = \dfrac{{k{{\left[ {0.012} \right]}^x}{{\left[ {0.035} \right]}^y}}}{{k{{\left[ {0.024} \right]}^x}{{\left[ {0.035} \right]}^y}}}$
$1 = {\left( {\dfrac{{0.012}}{{0.024}}} \right)^x}$
$1 = {\left( {\dfrac{1}{2}} \right)^x}$
${2^x} = 1$
${2^x} = {2^0}$ $\left[ {{a^0} = 1} \right]$
$x = 0$ $\left[ {{a^m} = {a^n}i.e.,m = n} \right]$
Now, expt.4 and 1 can be written as:
$0.80 = k{\left[ {0.012} \right]^x}{\left[ {0.070} \right]^y}$ and $0.10 = k{\left[ {0.012} \right]^x}{\left[ {0.035} \right]^y}$ .
We’ll divide both of them;
$\dfrac{{0.80}}{{0.10}} = \dfrac{{k{{\left[ {0.012} \right]}^x}{{\left[ {0.070} \right]}^y}}}{{k{{\left[ {0.012} \right]}^x}{{\left[ {0.035} \right]}^y}}}$
$8 = {\left( {\dfrac{{0.070}}{{0.035}}} \right)^y}$
$8 = {2^y}$
${2^3} = {2^y}$
$y = 3$
Therefore, the rate law is $R = k{\left[ A \right]^0}{\left[ B \right]^3}i.e.,R = k{\left[ B \right]^3}$

Note: Remember the formula of rate law of chemical reaction and formulas of exponents. Also, select both the Expt. No. in such a way so that one concentration gets cancelled while doing the calculations and k is constant for every particular reaction.