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}$$\\$ $=\dfrac{2.303}{1.15}*10^{-3}*0.2219\\ =444.38s\\ 444s(approx)$

6   Time required to decompose $SO_ 2 Cl_ 2$ to half its initial amount is $60$ minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

Solution :

For a $1^{ st}$ order reaction,$\\$ $t _{1/2} =\dfrac{0.693}{ k}$ $\\$ It is given that $ t_{1/2} = 60 $ min$\\$ $\therefore k=\dfrac{0.693}{t _{1/2}}$ $\\$ $=\dfrac{0.693}{ 60}$ $\\$ $= 0.01155$ min $^{ -1}$ $\\$ $= 1.155$ min $^{- 1}$ $\\$ or,$ k =1.925×10^{ -1} s^{ -1}$

7   What will be the effect of temperature on rate constant?

Solution :

The rate constant of a reaction is nearly doubled with a $10^ o$ rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,$\\$ $k = Ae ^{ Ea / Rt}$ $\\$ where,$\\$ A is the Arrhenius factor or the frequency factor$\\$ T is the temperature$\\$ R is the gas constant$\\$ Ea is the activation energy

8   The rate of the chemical reaction doubles for an increase of $10 K$ in absolute temperature from $298 K$. Calculate $ E a$ .

Solution :

Given $T_ 1 = 298 K$ $\\$ $\therefore T _2 = (298 + 10) K = 308 K$ $\\$ We also know that the rate of the reaction doubles when temperature is increased by $10K$ $\\$ Therefore, let us take the value of $k_ 1 = k$ and that of $k_ 2 = 2k$ $\\$ Also,$ R = 8.314 J K ^{- 1}$ mol ${-1}$ $\\$ Now substituting these values in the equation:$\\$ We get:$\\$ $\log\dfrac{k_2}{k_1}=\dfrac{E_a}{2.303R}[\dfrac{T_2-T_1}{T_1T_2}]$ $\\$ $\log\dfrac{2k}{k}=\dfrac{E_a}{2.303*8.314}[\dfrac{10}{298*308}]$ $\\$ $\log2=\dfrac{E_a}{2.303*8.314}[\dfrac{10}{298*308}]$ $\\$ $E_a=\dfrac{2.303*8.314*298*308*\log2}{10}$ $\\$ $=52897.78J$mol$^{ -1}$ $\\$ $=52.89kJ$ mol$^{ -1}$

9   The activation energy for the reaction $2HI _{(g)} \to H_ 2 + I_{ 2(g)}$ is $209.5 kJ$ mol $^{- 1}$ at $581 k$. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

Solution :

In the given case:$\\$ $Ea = 209.5 kJ ^{- 1} = 209500 J$ mol $^{ -1}$ $\\$ $T = 581 k$ $\\$ $R = 8.314Jk ^{- 1 } $ mol $^{- 1}$ $\\$ Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:$\\$ $x=e^{ Ea/Rt} $ $\\$ $\Rightarrow In= -Ea/RT$ $\\$ $\Rightarrow \log x=-\dfrac{Ea}{2.303RT}$ $\\$ $\Rightarrow \log x=\dfrac{209500 J mol^1}{2.303×8.314×Jk ^1 mol^{ -1} ×581} =18.8323 $ $\\$ Now $x =$ Antilog $(-18.8323)$ $\\$ $= 1.471x10 ^{-19}$

10   From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.$\\$ (i) $3 NO_{(g)} \to N_ 2 O _{(g)}$ Rate $ = k[NO]^ 2$ $\\$ (ii) $ H _2 O _{2 (aq)} + 3 I^{ -}_{ (aq)} + 2H ^+ \to 2 H_ 2 O _{(l)} + I_ {3^-}$ Rate$ = k[H_ 2 O_ 2 ] [I ^- ] $ $\\$ (iii) $CH _3 CHO_{(g)} \to CH_{ 4 (g)} + CO_{(g)}$ Rate $= k[CH _3 CHO]^{ 3/2}$ $\\$ (iv)$ C_ 2 H _5 Cl_{(g)}\to C _2 H_{ 4 (g)} + HCl_{(g)}$ Rate $= k[C _2 H _5 Cl]$

Solution :

(i) Given rate =$ k [NO]^ 2$ $\\$ Therefore, order of the reaction =$ 2$ $\\$ $k=\dfrac{\text{Rate}}{[No]^2}$ $\\$ Dimension of $k= \dfrac{\text{mol L}^{-1}s^{-1}}{(\text{mol L}^{-1})^2}$ $\\$ $=\dfrac{\text{mol L}^{-1}s^{-1}}{\text{mol$^2$ L$^{-2}$}}$ $\\$ $=L \ mol^{-1}s^{-1}$ $\\$ (ii) Given rate $k = [H_ 2 O _2 ] [I ^- ]$ $\\$ Therefore, order of the reaction $= 2$ $k=\dfrac{\text{Rate}}{ [H_ 2 O_ 2 ][ I ^- ]}$ $\\$ Dimension of $=\dfrac{\text{mol}^{-1}s^{-1}}{(\text{mol L}^{-1})(\text{mol L}^{-1})}$ $\\$ $=L^{-1} mol^{-1 } s^{-1}$ $\\$

(iii) Given rate =$ k [CH _3 CHO]^{ 3/2}$ $\\$ Therefore, order of reaction =$\dfrac{3}{2}$ $\\$ $k=\dfrac{\text{Rate}}{[CH_3CHO]^{3/2}}$ $\\$ Dimension of =$ \dfrac{\text{mol}^{-1}s^{-1}}{(\text{mol L}^{-1})^{3/2}}$ $\\$ $= \dfrac{\text{mol}^{-1}s^{-1}}{\text{mol }^{3/2} L^{3/2}}$$\\$ $=L^{1/2} mol^ {-1/2} s^{-1}$ $\\$ (iv) Given rate =$ k [C _2 H _5 Cl]$ $\\$ Therefore, order of the reaction $= 1 k = \dfrac{\text{Rate}}{[C_2H_5Cl]}$ $\\$ Dimension of = $ \dfrac{\text{mol L}^{-1}s^{-1}}{\text{mol L}^{-1}}$ $\\$ $=s^{-1}$