Question

Ethanal, dimerises in an alkaline solution according to the following equation.
$2C{{H}_{3}}CHO\to C{{H}_{3}}CH(OH)C{{H}_{2}}CHO$
The initial rate of this reaction was measured, starting with different concentrations of $C{{H}_{3}}CHO$ and $O{{H}^{-}}$. The following results were obtained.

$[C{{H}_{3}}CHO]/\text{mol d}{{\text{m}}^{-3}}$	$[O{{H}^{-}}]/\text{mol d}{{\text{m}}^{-3}}$	Initial rate of reaction (reactive value)
$0.10$	$0.015$	$1$
$0.20$	$0.015$	$2$
$0.40$	$0.030$	$3$

(i) Deduce the order of the reaction with respect to $C{{H}_{3}}CHO$.
(ii) Deduce the order of the reaction with respect to $O{{H}^{-}}$.
(iii) State the overall rate equation for this reaction. Rate $=$
(iv) State the unit for the rate constant, k.

Answer 1

To solve this question, you have to simply apply the concept of rate law or rate equation which generally gives an idea about the relationship between the rate of a chemical reaction and the concentration of the reactants in the reaction.

Complete step by step solution:
Rate laws or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the reactants concentration. The rate law is given as:
$rate=\text{ k }\\!\\![\\!\\!\text{ A}{{\text{ }\\!\\!]\\!\\!\text{ }}^{x}}{{\text{ }\\!\\![\\!\\!\text{ B }\\!\\!]\\!\\!\text{ }}^{y}}$ , where
k is the rate constant,
[A] and [B] are the molar concentrations of the reactants and
x and y represents the reaction order.
Given that, Ethanal, dimerises in an alkaline solution according to the following equation.
$2C{{H}_{3}}CHO\to C{{H}_{3}}CH(OH)C{{H}_{2}}CHO$
So, here the rate law can be written as:
$rate=\text{ k }\\!\\![\\!\\!\text{ C}{{\text{H}}_{3}}\text{CHO}{{\text{ }\\!\\!]\\!\\!\text{ }}^{x}}{{\text{ }\\!\\![\\!\\!\text{ O}{{\text{H}}^{-}}]}^{y}}$, where
$[C{{H}_{3}}CHO]$ represents the concentration of $C{{H}_{3}}CHO$ and
$[O{{H}^{-}}]$ represents the concentration of $O{{H}^{-}}$.
As per the information provided,
For first case, the rate law will be:
$1=\text{ k }\\!\\![\\!\\!\text{ 0}\text{.1}{{\text{ }\\!\\!]\\!\\!\text{ }}^{x}}{{\text{ }\\!\\![\\!\\!\text{ 0}\text{.015}]}^{y}}$ (i)
For second case, the rate law will be written as:
$2=\text{k }\\!\\![\\!\\!\text{ }\text{.2}{{\text{ }\\!\\!]\\!\\!\text{ }}^{x}}{{[0.015]}^{y}}$ (ii)
For third case, the rate law will be:
$8=\text{k }\\!\\![\\!\\!\text{ }\text{.4}{{\text{ }\\!\\!]\\!\\!\text{ }}^{x}}{{[0.030]}^{y}}$ (iii)
(i) Dividing equation (ii) from (i), we get:
$2={{\left[ \dfrac{0.2}{0.1} \right]}^{x}}$
Then, $2={{2}^{x}}$
So, $x=1$.
Thus, the order of the reaction with respect to $C{{H}_{3}}CHO$ is first order.
(ii) Dividing equation (iii) from (i), we get:
$8={{\left[ \dfrac{0.4}{0.1} \right]}^{x}}{{[2]}^{y}}$
Putting the value of x from above solution, we get:
$8={{\left[ 4 \right]}^{1}}{{[2]}^{y}}$
Then, $y=1$.
Thus, the order of the reaction with respect to $O{{H}^{-}}$ is first order.
(iii) The overall rate equation or rate law for this reaction is $rate=\text{ }\\!\\![\\!\\!\text{ C}{{\text{H}}_{3}}\text{CHO }\\!\\!]\\!\\!\text{ }\\!\\![\\!\\!\text{ O}{{\text{H}}^{-}}]$.
(iv) The unit for the rate constant is $mo{{l}^{-1}}d{{m}^{3}}{{s}^{-1}}$ as per any suitable time unit.

Note: It is important to note that; the rate of a reaction is a vital property because it says whether a reaction can occur during a lifetime. The rate law says about the concentration of the reactants and the order of the reaction.