Chemical Kinetics

Chemistry

NCERT

1   For the reaction $R \to P$, the concentration of a reactant changes from $0.03 M$ to $0.02M$ in $25$ minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Solution :

Average rate of reaction =$\dfrac{\Delta [R]}{\Delta t}$$\\$ $=-\dfrac{[R]_2-[R]_1}{t_2-t_1}$$\\$ $=-\dfrac{0.02-0.03}{25}M min^{-1}$$\\$ $=-\dfrac{(-0.01)}{25}M min^{-1}$$\\$ $=4*10^{-4}M min^{-1}$$\\$ $\dfrac{4*10^{-4}}{60}Ms^{-1}$$\\$ $=6.67*10^{-6}Ms^{-1}$

2   In a reaction, $2A \to$ Products, the concentration of $A$ decreases from $0.5$ mol $L^{-1}$ to $0.4 mol L^{-1}$ in $10$ minutes. Calculate the rate during this interval.

Solution :

Average rate $=-\dfrac{1}{2} \dfrac{\Delta [A]}{\Delta t}$$\\$ $=-\dfrac{1}{2}\dfrac{[A]_2-[A]_1}{t_2-t_1}$$\\$ $=-\dfrac{1}{2}*\dfrac{0.4-0.5}{10}$$\\$ $=-\dfrac{1}{2}*\dfrac{1-0.1}{10}$$\\$ $=0.005 mol L^{-1} min^{-1}=5*10^{-3}M min^{-1}$

3   For a reaction, $A + B \to $ Product; the rate law is given by $r = k [A]^{ 1/2} [B]^2$ . What is the order of the reaction?

Solution :

The order of the reaction $=\dfrac{1}{2}+2 =2\dfrac{1}{2}=2.5$

4   The conversion of molecules $X$ to $Y$ follows second order kinetics. If concentration of $x$ is increased to three times how will it affect the rate of formation of $Y$?

Solution :

The reaction $X\to Y$ follows second order kinetics hence the rate law equation will be$\\$ Rate $= kC^2 $, where $C = [x]$$\\$ If the concentration of $X$ is increases to three times, now $[x] = 3Cmol L^{-1}$$\\$ Now, the rate equation will be:$\\$ Rate $= k (3C)^2$$\\$ $= 9 (kC^2 )$$\\$ Thus the rate of reaction will become $9$ times$\\$ Hence, the rate of formation of $Y$ will increase by $9$ times

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

Solution :

Initial amount[R] $0 = 5 g$$\\$ Final concentration[R] $= 3 g$$\\$ Rate constant$ = 1.15× 10^{- 3} s ^{-1}$$\\$ We know that for a 1 st order reaction,$\\$ $t=\dfrac{2.303}{k}\log \dfrac{[R]_0}{R}$$\\$ $=\dfrac{2.303}{1.15 * 10^{-3}} \log \dfrac{5}{3}$$\\$ $=\dfrac{2.303}{1.15}*10^{-3}*0.2219\\ =444.38s\\ 444s(approx)$