Question

The rate of change in concentration of C in the reaction $2A+B\to 2C+3D$ was reported as $1.0\text{ mol litr}{\text{e}}^{-1}\text{ se}{\text{c}}^{-1}$. Calculate the reaction rate.
(A) $0.05\text{ mol litr}{\text{e}}^{-1}\text{ se}{\text{c}}^{-1}$
(B) $0.01\text{ mol litr}{\text{e}}^{-1}\text{ se}{\text{c}}^{-1}$
(C) $0.5\text{ mol litr}{\text{e}}^{-1}\text{ se}{\text{c}}^{-1}$
(D) $5\text{ mol litr}{\text{e}}^{-1}\text{ se}{\text{c}}^{-1}$

Answer 1

Hint: As we know that rate of a reaction is the speed at which the chemical reaction is taking place. During the reaction, concentration of the reactants decreases whereas the concentration of the product decreases. Also the change in concentration of the reactants and products affect the rate of the reaction. So we will use this definition to find out our answer.
Formula used:
For a reaction: $n_{1}A\to n_{2}B$
Rate of reaction = $-{\displaystyle \frac{1}{n_1}}{\displaystyle \frac{[dA]}{[dt]}}={\displaystyle \frac{1}{n_2}}{\displaystyle \frac{[dB]}{[dt]}}$

Complete answer:
We know that, in chemical reactions, reactants undergo the change (under suitable conditions) and finally yield the product. So rate of reaction tells us the speed at which this reaction takes place.
- During the reaction, concentration of the reactants decreases whereas the concentration of the product decreases. Also the change in concentration of the reactants and products affect the rate of the reaction. For the following reaction the rate of reaction in terms of concentration is shown below:-
$2A+B\to 2C+3D$
The rate of reaction = $-{\displaystyle \frac{1}{n_1}}{\displaystyle \frac{[dA]}{[dt]}}=-{\displaystyle \frac{1}{n_2}}{\displaystyle \frac{[dB]}{[dt]}}={\displaystyle \frac{1}{n_3}}{\displaystyle \frac{[dC]}{[dt]}}={\displaystyle \frac{1}{n_4}}{\displaystyle \frac{[dD]}{[dt]}}$
Here,
 $\begin{array}{rl}& n_1=2\\ & n_2=1\\ & n_3=2\\ & n_4=3\end{array}$
On replacing these values we get:-
Rate of reaction = $-{\displaystyle \frac{1}{2}}{\displaystyle \frac{[dA]}{[dt]}}=-{\displaystyle \frac{1}{1}}{\displaystyle \frac{[dB]}{[dt]}}={\displaystyle \frac{1}{2}}{\displaystyle \frac{[dC]}{[dt]}}={\displaystyle \frac{1}{3}}{\displaystyle \frac{[dD]}{[dt]}}$
-It is given that, rate of change in concentration of C ($[dC]/[dt]$) =$1.0\text{ mol litr}{\text{e}}^{-1}\text{ se}{\text{c}}^{-1}$. Hence by using this, we can calculate rate of reaction as follows:-
Rate of reaction = ${\displaystyle \frac{1}{2}}{\displaystyle \frac{[dC]}{[dt]}}$
Rate of reaction = ${\displaystyle \frac{1}{2}}\times 1.0\text{ mol litr}{\text{e}}^{-1}\text{ se}{\text{c}}^{-1}$
Rate of reaction = $0.5\text{ mol litr}{\text{e}}^{-1}\text{ se}{\text{c}}^{-1}$
-Hence the reaction rate for the given chemical reaction is: (C) $0.5\text{ mol litr}{\text{e}}^{-1}\text{ se}{\text{c}}^{-1}$

Note:
Always balance the reaction given in the question if it isn’t because the coefficients are useful in the calculation of reaction rate from the concentration.
-For a reaction: $n_{1}A\to n_{2}B$
Rate of reaction = $-{\displaystyle \frac{1}{n_1}}{\displaystyle \frac{[dA]}{[dt]}}={\displaystyle \frac{1}{n_2}}{\displaystyle \frac{[dB]}{[dt]}}$. Here the negative sign depicts that the concentration of A is decreasing and concentration of B is increasing.