Question

After how many seconds will the concentration of the reactant in a first order reaction be halved, if the rate constant is $\,\,1.155\times{10^{ - 3}}se{c^1}?\,\,$
A) 600
B) 100
C) 60
D) 10

Answer 1

A first-order reaction is dependent on the concentration of only one reactant. As such, a first-order reaction is sometimes known as a unimolecular reaction. While other reactants can exist, each will be zero-order, as the concentrations of these reactants do not affect the rate. Thus, the rate law for an elementary reaction that is first order with respect to a reactant $A$ is expressed as:
$\,\,{\text{r}} = - \dfrac{{{\text{d}}[{\text{A}}]}}{{{\text{dt}}}} = {\text{k}}[{\text{A}}]\,\,$
As usual, k is that the rate constant, and must have units of concentration/time; during this case it's units of $1/s$ . Usually, reaction rates decrease with time because reactant concentrations decrease as reactants are converted to products.

Complete step by step answer:
The speed law for a reaction may be a mathematical relationship between the reaction rate and also the concentrations of species in solution. Rate laws are expressed either as a differential rate law, describing the change in reactant or product concentrations as a function of your time, or as an integrated rate law, describing the particular concentrations of reactants or products as a function of your time.
The speed constant ($k$) of a rate law may be a constant of proportionality between the reaction rate and therefore the reactant concentration. The exponent to which a degree is raised in an exceedingly rate law indicates the reaction order, the degree to which the reaction rate depends on the Concentration of a specific reactant. Reaction rates generally increase when reactant concentrations are increased
The half-life of a reaction describes the time needed for half the reactant(s) to be depleted, which is that same because the half-life involved in nuclear decay, a first-order reaction. The formula for half lifetime of a reaction is as follows:
$\,\,t = \dfrac{{2.303}}{k}\log \frac{a}{{a - x}}\,\,$
Let Initial concentration $a = 100$
After half time, $x = 50$
Where t is the time, $a$ is the initial concentration and $x$ is the concentration at half life.
Substituting values in the formula, we get
$t = \dfrac{{2.303}}{{1.155 \times {{10}^{ - 3}}}}\log \dfrac{{100}}{{100 - 50}}$
On solving, we get
$t = 600\sec$
Hence the correct answer is option A.

Note: The proportionality constant ($k$) is named the speed constant, and its value is characteristic of the reaction and therefore the reaction conditions. A given reaction contains a particular rate constant value under a given set of conditions, like temperature, pressure, and solvent; varying the temperature or the solvent usually changes the worth of the speed constant. The numerical value of $k$, however, doesn't change because the reaction progresses under a given set of conditions.