Question

For an elementary reaction, $X(g)\to Y(g)+Z(g)$, the half life period is 10 min. In what period of time would concentration of X be reduced to 10% of original concentration?
(a)- 20 min
(b)- 33 min
(c)- 15 min
(d)- 25 min

Answer 1

Hint: The given reaction is $X(g)\to Y(g)+Z(g)$, so the rate of the reaction will be:
$Rate=k[X]$
So, it following the first-order kinetics, then find the rate constant by using the formula:
${{t}_{1/2}}=\dfrac{0.693}{k}$
Then to find the time of the completion of the reaction at specific concentration, use the formula:
$k=\dfrac{2.303}{t}\log \dfrac{a}{a-x}$
Where a is the initial concentration of the reactant and a-x is the final concentration of the reactant.

Complete answer:
In the question given above, we know the half life of the reaction is 10 min, and the given reaction is:
$X(g)\to Y(g)+Z(g)$
In this reaction only one reactant is present so, we write the rate of the reaction as:
$Rate=k[X]$
From this, we can see that this reaction will follow the first-order kinetics, and we can find the rate constant of the reaction by using the half-life of the reaction by using the formula:
${{t}_{1/2}}=\dfrac{0.693}{k}$
Putting, the value we can write:
$10=\dfrac{0.693}{k}$
$k=\dfrac{0.693}{10}=0.0693$
The k will be 0.0693
Now, to find the time of the completion of the reaction at specific concentration, i.e., at 10%, we can use the formula:
$k=\dfrac{2.303}{t}\log \dfrac{a}{a-x}$
Where a is the initial concentration of the reactant and a-x is the final concentration of the reactant.
So, a will be 100 and a – x will be 10, putting the values we get:
$t=\dfrac{2.303}{k}\log \dfrac{a}{a-x}$
$t=\dfrac{2.303}{0.0693}\log \dfrac{100}{10}$
$t=\dfrac{2.303}{0.0693}\log 10$
$t=33.23$
So, the time taken will be 33.23 min, and from the options, it is near to the 33 value.

Therefore, the correct answer is an option (b)- 33 min.

Note:
From the reactants, in the reaction, we can find the order of the reaction, when the value is not given. Half-life time is the time of the reaction when the concentration of the reactant becomes half of the initial concentration.