Courses
Courses for Kids
Free study material
Free LIVE classes
More

# The rate constant of reaction is $0.69 \times {10^{ - 1}}{\min ^{ - 1}}$ and the initial concentration is $0.2mol{L^{ - 1}}$. The half-life period is:A.$400\,\sec$B.$600\,\sec$C.$800\,\sec$D.$1200\,\sec$

Last updated date: 18th Mar 2023
Total views: 205.2k
Views today: 4.84k
Verified
205.2k+ views
Hint: In a chemical kinetics reaction rate constant or reaction rate coefficient, tell us about the rate of the direction and tell us about the direction of a chemical reaction. By looking at the reaction we can predict the direction of the reaction and also tell about the concentration of final products and leftover reactants.

We can say that when a substance A reacts with B to form C then we can say that rate of the reaction:
$aA + bB \to cC \\ r = k(T){[A]^m}{[B]^n} \\$
For a reaction as such, the rate of the reaction is directly proportional to molar concentrations of A and B given in the expression of rate. $k(T)$ is the reaction rate constant that depends completely on temperature. The exponents of A and B, constants are partial orders of reaction and not equal to the stoichiometric coefficient.
The reactions are further divided into different orders according to the number of molar coefficients of the elements on the left side of the reactions or the reactants of the product.
The orders in which they are divided are: -
First order
Second order
And so on…
The orders can be pre-decided by seeing the units of the reaction.
The first order reaction the reactants the reaction is dependent on is just one, the first order has two reactants, therefore, the coefficients of the first order are ${s^{ - 1}}$or any unit of time but in reverse proportion.
Now the rate of the reaction given is ${\min ^{ - 1}}$, therefore the order of the reaction is first.
As we know that the half life of a first order reaction or the time taken by the compound to reach the half concentration of what it was before starting the reaction.
We get the half time by the expression ${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{K}$
Now if we need the value of K, then we can take out of the reaction expression for a first order reaction:
$r = k[{A_ \circ }] \\ \Rightarrow k = \dfrac{r}{{[{A_ \circ }]}} \\$
Now putting the value of k in the half-life reaction
We get
${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k} \\ \Rightarrow {t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{r} \times [{A_ \circ }] \\$
$[{A_ \circ }]$ is the molar concentration of the compound, the initial concentration?
Taking out the value
${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{r} \times [{A_ \circ }] \\ \Rightarrow {t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{{0.69 \times {{10}^{ - 2}}}} \times 0.2 \\ \Rightarrow {t_{\dfrac{1}{2}}} = 1200{s^{ - 1}} \\$
The half time of the reaction $1200{s^{ - 1}}$.

Note:
The half-life of the half-life second order reaction is ${t_{\dfrac{1}{2}}} = \dfrac{1}{{k[{A_ \circ }]}}$. Second order of the reaction is defined as that reaction in which the sum of the exponents is greater than $2$. Many important biological reactions, such as the formation of the double-stranded DNA.