
A first order reaction has a rate constant value of $0.0051\,{{\min }^{-1}}$ .If we begin with $0.10\,M$ concentration of the reactant, how much of the reaction will remain after $300 hours$.
Answer
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Hint: The concept to be used in this question is rate analysis of first order reaction.the rate in this reaction varies exponentially with the concentration of the reaction.
- The rate constant defines the relation between the rate of reaction and the concentration of the reactants.
- Convert all the given quantities in the same units and here it will be minutes.
Use the standard first order equation to find the reactant left after $300 hours$.
Complete step by step solution:
The first order rate equation which connects the rate constant and concentration of reactant with time is:
$k=\dfrac{2.303}{t}{{\log }_{e}}\dfrac{{{A}_{o}}}{A}$, where $k$ is rate constant , $t$ is time, $A$ is initial concentration and ${{A}_{o}}$ is left over concentration after time $t$.
Given values,$k$ = $0.0051\,{{\min }^{-1}}$ $t$ = $3\,hours$ = $3\times 60\,minutes$
$A$ = $0.10\,M$
As the quantities are in the same units now,we will use the first order rate equation to find the concentration after the given time.
Putting the values:
$\Rightarrow 0.0051=\dfrac{2.303}{3\times 60}{{\log }_{e}}\dfrac{0.10}{A}$
$\Rightarrow {{\log }_{e}}\dfrac{0.10}{A} = \dfrac{0.0051\times 180}{2.303}$
$\Rightarrow {{\log }_{e}}\dfrac{0.10}{A} = 0.3986$
Taking antilog to the base e on both sides,
$\Rightarrow \dfrac{0.10}{A} = 2.503$
$\Rightarrow A = 0.039M$
Hence the concentration of the reactant after $3$ hours is $0.039\,M$.
Note: The reactions always follow a particular order when they proceed.there is zeroth order reaction,first order reaction second order reaction and nth order reaction.
- Half life of a reaction is defined as the time after which the concentration is reduced to half of its initial value.
- We have to take the antilog to the natural base that is e while solving the equation.
Also ,for zeroth order reaction, half life is independent of its initial concentration.
- The rate constant defines the relation between the rate of reaction and the concentration of the reactants.
- Convert all the given quantities in the same units and here it will be minutes.
Use the standard first order equation to find the reactant left after $300 hours$.
Complete step by step solution:
The first order rate equation which connects the rate constant and concentration of reactant with time is:
$k=\dfrac{2.303}{t}{{\log }_{e}}\dfrac{{{A}_{o}}}{A}$, where $k$ is rate constant , $t$ is time, $A$ is initial concentration and ${{A}_{o}}$ is left over concentration after time $t$.
Given values,$k$ = $0.0051\,{{\min }^{-1}}$ $t$ = $3\,hours$ = $3\times 60\,minutes$
$A$ = $0.10\,M$
As the quantities are in the same units now,we will use the first order rate equation to find the concentration after the given time.
Putting the values:
$\Rightarrow 0.0051=\dfrac{2.303}{3\times 60}{{\log }_{e}}\dfrac{0.10}{A}$
$\Rightarrow {{\log }_{e}}\dfrac{0.10}{A} = \dfrac{0.0051\times 180}{2.303}$
$\Rightarrow {{\log }_{e}}\dfrac{0.10}{A} = 0.3986$
Taking antilog to the base e on both sides,
$\Rightarrow \dfrac{0.10}{A} = 2.503$
$\Rightarrow A = 0.039M$
Hence the concentration of the reactant after $3$ hours is $0.039\,M$.
Note: The reactions always follow a particular order when they proceed.there is zeroth order reaction,first order reaction second order reaction and nth order reaction.
- Half life of a reaction is defined as the time after which the concentration is reduced to half of its initial value.
- We have to take the antilog to the natural base that is e while solving the equation.
Also ,for zeroth order reaction, half life is independent of its initial concentration.
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