Question
Question: A first order reaction has a rate constant value of \(0.0051\,{{\min }^{-1}}\) .If we begin with \(0...
A first order reaction has a rate constant value of 0.0051min−1 .If we begin with 0.10M concentration of the reactant, how much of the reaction will remain after 300hours.
Solution
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 300hours.
Complete step by step solution:
The first order rate equation which connects the rate constant and concentration of reactant with time is:
k=t2.303logeAAo, where k is rate constant , t is time, A is initial concentration and Ao is left over concentration after time t.
Given values,k = 0.0051min−1 t = 3hours = 3×60minutes
A = 0.10M
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:
⇒0.0051=3×602.303logeA0.10
⇒logeA0.10=2.3030.0051×180
⇒logeA0.10=0.3986
Taking antilog to the base e on both sides,
⇒A0.10=2.503
⇒A=0.039M
Hence the concentration of the reactant after 3 hours is 0.039M.
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.