Question

1) For the reaction \[R \to P\], the concentration of the reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
(2) In a reaction, \[2A \to P\] products, the concentration of A decreases from 0.5 to 0.4 mol \[{L^{ - 1}}\]in 10 minutes. Calculate the rate during this interval.

Answer 1

Hint: The average rate of reaction is the ratio of the change in concentration of the reactant or the product in the chemical reaction to the time interval. The negative value and positive value of the average rate depend on change in the concentration.

Complete step by step answer:
(1) Change in concentration from 0.03M to 0.02M in 25 minutes.
(2) Decrease in concentration from 0.5 to 0.4 mol \[{L^{ - 1}}\] in 10 minutes.
(1) The average rate of reaction, is defined as the ratio of the concentration change of the reactant or the product of the chemical reaction to the time interval.
The mathematical expression used to calculate the average rate is shown below.
\[Average\;rate = \dfrac{{\Delta X}}{{\Delta t}}\]
Where,
\[\Delta X\] is the change in concentration of reactant or product
\[\Delta t\] is the time interval
The given reaction is shown below.
\[R \to P\]
Here, the reactant converts into a product.
Substitute the values in the above equation.
\[ \Rightarrow Average\;rate = \dfrac{{0.03 - 0.02}}{{25}}\]
\[ \Rightarrow Average\;rate = \dfrac{{0.01}}{{25}}\]
\[ \Rightarrow Average\;rate = 4 \times {10^{ - 4}}M/\min \]
\[ \Rightarrow Average\;rate = \dfrac{{0.03 - 0.02}}{{25 \times 60}}\]
\[ \Rightarrow Average\;rate = \dfrac{{0.01}}{{25 \times 60}}\]
\[ \Rightarrow Average\;rate = 6.7 \times {10^{ - 6}}M/sec\]
Therefore, the average rate of reaction in minutes is \[4 \times {10^{ - 4}}M/\min \] and in seconds \[6.7 \times {10^{ - 6}}M/sec\].
(2) The rate is calculated by the formula as shown below.
\[\;rate = - \dfrac{1}{2}\left( {\dfrac{{\Delta A}}{{\Delta t}}} \right)\]
Where,
\[\Delta A\] is the decrease in the concentration of A.
\[\Delta t\]is the time interval.
The given reaction is shown below.
\[2A \to P\]
Here reactant 2A changes to product P.
Substitute the values in the above equation.
\[\; \Rightarrow rate = - \dfrac{1}{2}\left( {\dfrac{{0.4 - 0.5}}{{10}}} \right)\]
\[\; \Rightarrow rate = - \dfrac{1}{2}\left( {\dfrac{{ - 0.1}}{{10}}} \right)\]
\[\; \Rightarrow rate = 0.005mol{L^{ - 1}}{\min ^{ - 1}}\]
\[\; \Rightarrow rate = 5 \times {10^{ - 3}}mol{L^{ - 1}}{\min ^{ - 1}}\]
Therefore, the rate is \[5 \times {10^{ - 3}}mol{L^{ - 1}}{\min ^{ - 1}}\].

Note: The average rate of reaction can either be positive or negative. It is said to be positive when there is an increase in the rate of concentration of product. It is said to be negative when there is a decrease in the rate of concentration of the reactant.