
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
562.8k+ views
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

