Question

Out of $300{\text{ }}g$substance (decomposes as per ${1^{st}}$ order). How much will remain after$18{\text{ }}hrs$?
(${t_{\dfrac{1}{2}}} = {\text{ }}3{\text{ }}hrs$.)

$$A.\;\;\;\;\;4.6{\text{ }}gm \\\ B.\;\;\;\;\;5.6{\text{ }}gm \\\ C.\;\;\;\;\;9.2{\text{ }}gm \\\ D.\;\;\;\;\;6.4{\text{ }}gm \\\$$

Answer 1

${1^{st}}$ order reactions are the reactions which the reaction rate depend upon linearly on the concentration of only one reactant. And the order of these reactions is one. The rate of the reaction for first order is $K = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{{{{[A]}_0}}}{{{{[A]}_t}}}$.

Complete answer: The question above is of first order reaction. The first order reactions are those in which the rate of the reaction depends upon the concentration of only one reactant. It is given that out of $300{\text{ }}g$ a certain amount of substance has been decomposed at a time period of 18 hrs. We need to find the amount of substance remained after $18{\text{ }}hrs$. Also given that ${t_{\dfrac{1}{2}}} = {\text{ }}3{\text{ }}hrs$.
For solving this question we will use the formula of first order reaction:
$\Rightarrow$ $K = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{{{{[A]}_0}}}{{{{[A]}_t}}}$
Let us see what all values we have, given that ${[A]_0} = 300{\text{ }}g$,$t{\text{ }} = {\text{ }}18{\text{ }}hrs$, ${[A]_t}$ we have to calculate, even value of $K$ is not given but we have the value of ${t_{\dfrac{1}{2}}} = {\text{ }}3{\text{ }}hrs$ so by using the equation of half-life we can calculate what is ‘$K$’.
The equation for half-life is:${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{K}$ For finding $K$ we can rearrange the formula as $K = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$
$\Rightarrow$ $K = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$ ${t_{\dfrac{1}{2}}} = 3hrs.$
$K = \dfrac{{0.693}}{3}$
$\Rightarrow$ $K = 0.231h{r^{ - 1}}$
Now we can substitute the value of $K$in the first order reaction to find out the value of ${[A]_t}$
Therefore $K = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{{{{[A]}_0}}}{{{{[A]}_t}}}$ we will rearrange this equation as ${\log _{10}}\dfrac{{{{[A]}_0}}}{{{{[A]}_t}}} = \dfrac{{K \times t}}{{2.303}}$
$\Rightarrow{\log _{10}}\dfrac{{{{[A]}_0}}}{{{{[A]}_t}}} = \dfrac{{K \times t}}{{2.303}}$ Substituting the values of ‘$t$’ and ‘$K$’we have
$\Rightarrow{\log _{10}}\dfrac{{{{[A]}_0}}}{{{{[A]}_t}}} = \dfrac{{0.231 \times 18}}{{2.303}}$
$\Rightarrow{\log _{10}}\dfrac{{{{[A]}_0}}}{{{{[A]}_t}}} = {\text{1}}{\text{.8055}}$
Now we will shift the log on another side, when you shift a log from one side to another it becomes antilog. And antilog is ${10^x}$ where x is the number for which antilog is to be found out
Thus $\Rightarrow$ $\dfrac{{{{[A]}_0}}}{{{{[A]}_t}}} = anti\log ({\text{1}}{\text{.8055)}}$
$\dfrac{{{{[A]}_0}}}{{{{[A]}_t}}} = {10^{{\text{1}}{\text{.8055}}}}$
$\Rightarrow\dfrac{{{{[A]}_0}}}{{{{[A]}_t}}} = 63.8956$ It is given that ${[A]_0}$= 300 g
So $\dfrac{{300}}{{{{[A]}_t}}} = 63.8956$
${[A]_t} = \dfrac{{300}}{{63.8956}}$
Hence $\Rightarrow$ ${[A]_t} = 4.6g$
Therefore the correct option is $A.\;\;\;\;\;4.6{\text{ }}gm$.

Note:
The unit of first order reaction is${S^{ - 1}}$. Hydrolysis of aspirin is one of the examples of first order reactions. Other examples are absorption, distribution, elimination rates etc. there are two more types of order of reaction among which zero order, second order and pseudo order are mostly considered along with first order.