Question

The results given in the below table were obtained during kinetic studies of the following reaction: \[{\rm{2A + B}} \to {\rm{C + D}}\]

Experiment	[A] / molL-1	[B] / molL-1	Initial rate/molL-1min-1
I	0.1	0.1	\[{\rm{6}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{ - 3}}\]
II	0.1	0.2	\[{\rm{2}}{\rm{.40}} \times {\rm{1}}{{\rm{0}}^{ - 2}}\]
III	0.2	0.1	\[{\rm{1}}{\rm{.20}} \times {\rm{1}}{{\rm{0}}^{ - 2}}\]
IV	X	0.2	\[{\rm{7}}{\rm{.20}} \times {\rm{1}}{{\rm{0}}^{ - 2}}\]
V	0.3	Y	\[2.88 \times {\rm{1}}{{\rm{0}}^{ - 1}}\]

X and Y in the given table are respectively:
1) \[0.4,\,0.4\]
2) \[0.3,\,0.4\]
3) \[0.4,\,0.3\]
4) \[0.3,\,0.3\]

Answer 1

Hint: The order of a reaction may be defined as the sum of the exponents to which the concentration terms in the rate law are raised to express the observed rate of reaction.

From general equation of the type: \[{\rm{aA}}\,\,{\rm{ + }}\,\,{\rm{bB}} \to {\rm{products}}\], the rate expression can be written as, \[{\rm{rate}}\,\, = \,{{\rm{k}}_{\rm{r}}}{{\rm{[A]}}^{\rm{a}}}{{\rm{[B]}}^{\rm{b}}}\].
\[{\rm{a}}\] and \[{\rm{b}}\]are called the orders of the reaction with respect to \[{\rm{A}}\]and \[{\rm{B}}\]. Depending on whether \[{\rm{(a + b)}}\]is equal to \[{\rm{0,1,2, or 3}}\], the reactions are said to be of zero order, first order, second order, and third order respectively.

Complete Step by Step Solution:
Rate equation may be defined as the relationship between the concentrations of the reactants and the observed reaction rate. Each reaction has its own rate equation which may be determined experimentally by altering the concentration of the reactants and then measuring the change in rate.
Consider a general reaction of the type: \[{\rm{aA}}\,\,{\rm{ + }}\,\,{\rm{bB}} \to {\rm{products}}\]

The reaction rate is usually proportional to the concentrations of the reactants \[{\rm{[A]}}\] and \[{\rm{[B]}}\] which are raised to powers \[{\rm{a}}\] and \[{\rm{b}}\] respectively. The rate expression can, then be written as:
\[{\rm{rate}}\,\, = \,{{\rm{k}}_{\rm{r}}}{{\rm{[A]}}^{\rm{a}}}{{\rm{[B]}}^{\rm{b}}}\], where \[{{\rm{k}}_{\rm{r}}} = \]rate constant.
As per given data,
The equation is as: \[{\rm{2A + B}} \to {\rm{C + D}}\]
Suppose the rate equation is, \[{\rm{rate}}\,\, = \,{\rm{k[A}}{{\rm{]}}^{\rm{a}}}{{\rm{[B]}}^{\rm{b}}}\]
where, \[{\rm{a}}\] and \[{\rm{b}}\]are the orders with respect to \[{\rm{[A]}}\] and \[{\rm{[B]}}\]

Substituting the given values in the rate equation we get,
\[{\rm{(rate) I 6}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{ - 3}} = {\rm{k (0}}{\rm{.1}}{{\rm{)}}^{\rm{a}}}{{\rm{(0}}{\rm{.1)}}^{\rm{b}}}\]
\[{\rm{(rate) II 2}}{\rm{.40}} \times {\rm{1}}{{\rm{0}}^{ - 2}} = {\rm{k (0}}{\rm{.1}}{{\rm{)}}^{\rm{a}}}{{\rm{(0}}{\rm{.2)}}^{\rm{b}}}\]
\[{\rm{(rate) III 1}}{\rm{.20}} \times {\rm{1}}{{\rm{0}}^{ - 2}} = {\rm{k (0}}{\rm{.2}}{{\rm{)}}^{\rm{a}}}{{\rm{(0}}{\rm{.1)}}^{\rm{b}}}\]
Dividing experiment (I) by experiment (II) we get,
$\dfrac{{{\rm{exp}}\,\,{\rm{I}}}}{{{\rm{exp}}\,\,{\rm{II}}}} = \dfrac{{{\rm{6}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{ - 3}}}}{{{\rm{2}}{\rm{.40}} \times {\rm{1}}{{\rm{0}}^{ - 2}}}} = \dfrac{{\not{{\rm{k}}}{\rm{ }}\not{{{{{\rm{(0}}{\rm{.1)}}}^{\rm{a}}}}}{{{\rm{(0}}{\rm{.1)}}}^{\rm{b}}}}}{{\not{{\rm{k}}}{\rm{ }}\not{{{{{\rm{(0}}{\rm{.1)}}}^{\rm{a}}}}}{{{\rm{(0}}{\rm{.2)}}}^{\rm{b}}}}}\\ $
$\Rightarrow \dfrac{{{\rm{exp}}\,\,{\rm{I}}}}{{{\rm{exp}}\,\,{\rm{II}}}} = \dfrac{{\rm{1}}}{{\rm{4}}}{\rm{ = (}}\dfrac{{\rm{1}}}{{\rm{2}}}{{\rm{)}}^{\rm{b}}}\\ $
$\Rightarrow \dfrac{{{\rm{exp}}\,\,{\rm{I}}}}{{{\rm{exp}}\,\,{\rm{II}}}} = {(\dfrac{{\rm{1}}}{2}{\rm{)}}^2}{\rm{ = (}}\dfrac{{\rm{1}}}{{\rm{2}}}{{\rm{)}}^{\rm{b}}}\\ $
$\Rightarrow {\rm{b = 2}}$
So, order with respect to \[{\rm{[B]}}\] is \[{\rm{2}}\].

Also, dividing experiment (I) by experiment (III) we get,
$\dfrac{{{\rm{exp}}\,\,{\rm{I}}}}{{{\rm{exp}}\,\,{\rm{III}}}} = \dfrac{{{\rm{6}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{ - 3}}}}{{{\rm{1}}{\rm{.20}} \times {\rm{1}}{{\rm{0}}^{ - 2}}}} = \dfrac{{\not{{\rm{k}}}{\rm{ (0}}{\rm{.1}}{{\rm{)}}^{\rm{a}}}\not{{{{{\rm{(0}}{\rm{.1)}}}^{\rm{b}}}}}}}{{\not{{\rm{k}}}{\rm{ (0}}{\rm{.2}}{{\rm{)}}^{\rm{a}}}\not{{{{{\rm{(0}}{\rm{.1)}}}^{\rm{b}}}}}}}\\ $
$\Rightarrow \dfrac{{{\rm{exp}}\,\,{\rm{I}}}}{{{\rm{exp}}\,\,{\rm{III}}}} = \dfrac{{\rm{1}}}{2}{\rm{ = (}}\dfrac{{\rm{1}}}{{\rm{2}}}{{\rm{)}}^{\rm{a}}}\\$
$ \Rightarrow \dfrac{{{\rm{exp}}\,\,{\rm{I}}}}{{{\rm{exp}}\,\,{\rm{III}}}} = {(\dfrac{{\rm{1}}}{2}{\rm{)}}^1}{\rm{ = (}}\dfrac{{\rm{1}}}{{\rm{2}}}{{\rm{)}}^{\rm{a}}}\\ $
$\Rightarrow {\rm{a = 1}}$
So, order with respect to \[{\rm{[A]}}\]is \[{\rm{1}}\].

Again, dividing experiment (I) by experiment (IV) we get,
$\dfrac{{{\rm{exp}}\,\,{\rm{I}}}}{{{\rm{exp}}\,\,{\rm{IV}}}} = \dfrac{{{\rm{6}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{ - 3}}}}{{{\rm{7}}{\rm{.20}} \times {\rm{1}}{{\rm{0}}^{ - 2}}}} = \dfrac{{\not{{\rm{k}}}{\rm{ (0}}{\rm{.1}}{{\rm{)}}^1}{{{\rm{(0}}{\rm{.1)}}}^2}}}{{\not{{\rm{k}}}{\rm{ (X}}{{\rm{)}}^1}{{(0.2)}^2}}}\\$
$ \Rightarrow \dfrac{{{\rm{exp}}\,\,{\rm{I}}}}{{{\rm{exp}}\,\,{\rm{IV}}}} = \dfrac{{\rm{1}}}{{12}}{\rm{ = (}}\dfrac{{0.1}}{{\rm{X}}}{\rm{)(}}\dfrac{{0.01}}{{0.04}}{\rm{)}}\\ $
$\Rightarrow \dfrac{{\rm{1}}}{{12}}{\rm{ = (}}\dfrac{{0.1}}{{\rm{X}}}{\rm{)(}}\dfrac{{0.01}}{{0.04}}{\rm{)}}\\ $
$\Rightarrow \dfrac{{\rm{1}}}{{12}}{\rm{ = }}\dfrac{{0.001}}{{0.04{\rm{X}}}}\\ $
$\Rightarrow {\rm{X}} = \dfrac{{12 \times 0.001}}{{1 \times 0.04}}\\$
$ \Rightarrow {\rm{X}} = \dfrac{{0.012}}{{0.04}}\\$
$ \Rightarrow {\rm{X}} = 0.3$
Hence, the value of X is found to be \[{\rm{0}}{\rm{.3}}\]

Finally, dividing experiment (I) by experiment (V) we get,
$\dfrac{{{\rm{exp}}\,\,{\rm{I}}}}{{{\rm{exp}}\,\,{\rm{V}}}} = \dfrac{{{\rm{6}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{ - 3}}}}{{2.88 \times {\rm{1}}{{\rm{0}}^{ - 1}}}} = \dfrac{{\not{{\rm{k}}}{\rm{ (0}}{\rm{.1}}{{\rm{)}}^1}{{{\rm{(0}}{\rm{.1)}}}^2}}}{{\not{{\rm{k}}}{\rm{ (0}}{\rm{.3}}{{\rm{)}}^1}{{({\rm{Y}})}^2}}}\\$
$ \Rightarrow \dfrac{{{\rm{exp}}\,\,{\rm{I}}}}{{{\rm{exp}}\,\,{\rm{V}}}} = \dfrac{{\rm{1}}}{{48}}{\rm{ = (}}\dfrac{{0.1}}{{0.3}}{\rm{)(}}\dfrac{{0.01}}{{{{\rm{Y}}^{\rm{2}}}}}{\rm{)}}\\$
$ \Rightarrow \dfrac{{\rm{1}}}{{48}}{\rm{ = (}}\dfrac{{0.1}}{{0.3}}{\rm{)(}}\dfrac{{0.01}}{{{{\rm{Y}}^{\rm{2}}}}}{\rm{)}}\\$
$ \Rightarrow \dfrac{{\rm{1}}}{{48}}{\rm{ = }}\dfrac{{0.001}}{{0.3{{\rm{Y}}^{\rm{2}}}}}\\ $
$\Rightarrow {{\rm{Y}}^{\rm{2}}} = \dfrac{{48 \times 0.001}}{{1 \times 0.3}}\\ $
$\Rightarrow {{\rm{Y}}^{\rm{2}}} = \dfrac{{0.048}}{{0.3}}\\ $
$\Rightarrow {{\rm{Y}}^{\rm{2}}} = 0.16\\\Rightarrow {{\rm{Y}}^{\rm{2}}} = {(0.4)^2}\\ $
$\Rightarrow {\rm{Y}} = 0.4$
Hence, the value of Y is found to be \[0.4\]
Therefore, option (2) is correct.

Note: The order with respect to \[{\rm{[A]}}\]is \[{\rm{1}}\]and the order with respect to \[{\rm{[B]}}\] is \[{\rm{2}}\]. The overall order for the reaction will be \[(1 + 2)\, = 3\]. This implies that the reaction is of third order overall. Also, the units of rate constant (\[{\rm{k}}\]) for the given reaction can be calculated.
${\rm{rate}}\,\, = \,{\rm{k[A}}{{\rm{]}}^1}{{\rm{[B]}}^2}\\ $
$\Rightarrow {\rm{k}} = \dfrac{{{\rm{rate}}}}{{{{{\rm{[A]}}}^{\rm{1}}}{{{\rm{[B]}}}^{\rm{2}}}}}\\ \Rightarrow {\rm{k}} = \dfrac{{{\rm{mol}}{{\rm{L}}^{{\rm{ - 1}}}}{\rm{mi}}{{\rm{n}}^{{\rm{ - 1}}}}}}{{{{{\rm{(mol}}{{\rm{L}}^{{\rm{ - 1}}}}{\rm{)}}}^{\rm{1}}}{{{\rm{(mol}}{{\rm{L}}^{{\rm{ - 1}}}}{\rm{)}}}^{\rm{2}}}}}\\$
$ \Rightarrow {\rm{k}} = {\rm{mo}}{{\rm{l}}^{ - 2}}{{\rm{L}}^2}{\rm{mi}}{{\rm{n}}^{{\rm{ - 1}}}}$
The units of rate constant is found to be \[{\rm{mo}}{{\rm{l}}^{ - 2}}{{\rm{L}}^2}{\rm{mi}}{{\rm{n}}^{{\rm{ - 1}}}}\].