Question

An organic compound undergoes first order decomposition. The time taken for its decomposition is $ \dfrac{1}{8} $ and $ \dfrac{1}{{10}} $ of its initial concentration are $ {t_{\dfrac{1}{8}}} $ $ {t_{\dfrac{1}{8}}} $ and $ {t_{\dfrac{1}{{10}}}} $ respectively. The value of $ \dfrac{{\left[ {{t_{\dfrac{1}{8}}}} \right]}}{{\left[ {{t_{\dfrac{1}{{10}}}}} \right]}} \times 10 $ is ? $ \left( {{{\log }_{10}}2 = 0.3} \right) $

Answer 1

Hint :First order reaction is a unimolecular reaction that contains only one reactant. The rate of reaction varies with the change in the concentration of only one reactant. Therefore, the order of the reaction is equal to one.
 $ t = \dfrac{{2.303}}{k}\log \left( {\dfrac{a}{{a - x}}} \right) $
Where, $ t $ is the time,
 $ k $ is the rate constant,
 $ a $ is the initial concentration,
 $ a - x $ is the concentration at time $ t $

Complete Step By Step Answer:
In the question it is given that the initial concentration at $ {t_{\dfrac{1}{8}}} $ and $ {t_{\dfrac{1}{{10}}}} $ is $ 1 $ $ $
The concentration at $ {t_{\dfrac{1}{8}}} $ is $ \dfrac{1}{8} $
The concentration at $ {t_{\dfrac{1}{{10}}}} $ is $ \dfrac{1}{{10}} $
 $ t = \dfrac{{2.303}}{k}\log \left( {\dfrac{a}{{a - x}}} \right) $
Now substituting the value in the above formula we get
A) $ {t_{\dfrac{1}{8}}} = \dfrac{{2.303}}{k}\log \left( {\dfrac{1}{{\dfrac{1}{8}}}} \right) $
 $ \Rightarrow {t_{\dfrac{1}{8}}} = \dfrac{{2.303}}{k}\log \left( 8 \right) $
 $ \Rightarrow {t_{\dfrac{1}{8}}} = \dfrac{{2.303}}{k}\left( {0.903} \right) $
 $ \Rightarrow {t_{\dfrac{1}{8}}} = \dfrac{{2.079}}{k} $
 $ {t_{\dfrac{1}{{10}}}} = \dfrac{{2.303}}{k}\log \left( {\dfrac{1}{{\dfrac{1}{{10}}}}} \right) $
 $ \Rightarrow {t_{\dfrac{1}{{10}}}} = \dfrac{{2.303}}{k}\log \left( {10} \right) $
 $ \Rightarrow {t_{\dfrac{1}{{10}}}} = \dfrac{{2.303}}{k} $
The value of $ \dfrac{{\left[ {{t_{\dfrac{1}{8}}}} \right]}}{{\left[ {{t_{\dfrac{1}{{10}}}}} \right]}} \times 10 = \dfrac{{\dfrac{{2.079}}{k}}}{{\dfrac{{2.303}}{k}}} \times 10 $
 $ \Rightarrow \dfrac{{\left[ {{t_{\dfrac{1}{8}}}} \right]}}{{\left[ {{t_{\dfrac{1}{{10}}}}} \right]}} \times 10 = 9 $

Additional Information:
The rate of chemical reaction is defined as the speed at which a chemical reaction is moving forward.
The factors that affect the rate of a reaction are:
Surface area – is directly proportional to the rate of reaction. Increasing the surface area of a reactant increases the rate of reaction.
Temperature – is also directly proportional to the rate of reaction.
Reactant concentration: Increasing the concentration of a reactant increases the rate of reaction.

Note :
First order reaction is defined as a reaction that only depends on the concentration of one reactant.
In this question, we have used the formula of first order reaction to calculate the value of $ \dfrac{{\left[ {{t_{\dfrac{1}{8}}}} \right]}}{{\left[ {{t_{\dfrac{1}{{10}}}}} \right]}} \times 10 $ .
if the reactant at first order reaction is doubled, then the reaction rate is also doubled.