Question

A first order reaction has a rate constant $1.15 \times {10^{ - 3}}{s^{ - 1}}$. How long will $5g$ of this reactant take to reduce to $4g$ ?

Answer 1

It is important to determine the rate of the reaction which depends on the concentration of the reactants and time taken by the reaction. The rate at a particular instant is known as the instantaneous rate. We calculate the time taken by using the formula for the first order reaction which is a linear reaction which depends on only one reactant.
Formula used:
$k = \dfrac{{2.303}}{t}\log \dfrac{{\left[ {{R_0}} \right]}}{{\left[ R \right]}}$

Complete step by step answer:
We know that it is important to understand the extent, feasibility and application of a reaction but it is also important to determine the rate of reaction and the factors that affect the rate of the reaction. The rate of reaction depends on the concentration of the reactants and the time taken for the reaction as the rate of reaction decreases with increase in time as the concentration of the reactants decreases with time. The rate of reaction is calculated as $Rate = k{\left[ A \right]^x}{\left[ B \right]^y}$. The sum of the powers of the concentration of the reactants in the above expression gives the order of the chemical reaction.
The first order reaction is a concentration dependent reaction. It is a linear reaction in which the rate of reaction depends on only one concentration. The unstable radioactive decay of the nuclei takes place by the first order reaction.
For the above question we use the formula for calculating the rate of first order reaction
$k = \dfrac{{2.303}}{t}\log \dfrac{{\left[ {{R_0}} \right]}}{{\left[ R \right]}}$
We know that,
$t = \dfrac{{2.303}}{k}\log \dfrac{{\left[ {{R_0}} \right]}}{{\left[ R \right]}}$
It is given that $k = 1.15 \times {10^{ - 3}}{s^{ - 1}}$
${R_0} = 5g$ and $R = 4g$
$\Rightarrow t = \dfrac{{2.303}}{{1.15 \times {{10}^{ - 3}}}}\log \left( {\dfrac{5}{4}} \right)$
$\Rightarrow t = 2 \times {10^3}\log (1.25)$
$\Rightarrow t = 2 \times {10^3} \times 0.096$
$\Rightarrow t = 192s$

Thus, the first order reaction will take $192s$ to reduce from $5g$ to $4g$ with a rate constant of $1.15 \times {10^{ - 3}}$.

Note:
The second order reaction is sometimes treated like a first order reaction because the second order reaction is difficult to execute due to the simultaneous measurement of the concentration of two reactants which may result in various complications and high expenses of the reaction. Thus a second order reaction is sometimes treated as a pseudo-first order reaction.