Name: Sequential First Order Reactions
Uploaded: 2019-07-19
Duration: 9 min 41 s
Description: Sequential reactions consist of linked reactions in which the product of the first reaction becomes the substrate of a second reaction. Let us learn more abo...

Question

For a consecutive first-order reaction $A \overset{k_{1}}{⟶} B \overset{k_{2}}{⟶} C$ the values of $k_{1}$ and $k_{2}$ are $45 {hr}^{- 1}$ and $15 {hr}^{- 1}$ , respectively. If the reaction is carried out with pure $A$ at a concentration of $1.0 mol {dm}^{- 3}$ , what will be the composition of the reacting system after a time interval of $10$ minutes?

Accepted Answer

The given consecutive reaction is,

$A \overset{k_{1}}{⟶} B \overset{k_{2}}{⟶} C$

Where, $k_{1} = 45 {hr}^{- 1}$ and $k_{2} = 15 {hr}^{- 1}$ are the rate constants.

In the above consecutive reaction, two reactions are taking place.

Rate of first reaction $=$ $k_{1} [A]$

Rate of first reaction $=$ $k_{2} [B]$

Compound $A$ is consumed in the first reaction to form $B$ . So, the rate of consumption of $A$ will be equal to the rate of the first reaction.

Compound $B$ is formed from first reaction and at the same time is consumed in second reaction. Overall, the rate of formation of $B$ will be equal to the difference in the rate of first and rate of second reaction.

Compound $C$ is formed in the second reaction. So, the rate of formation of $C$ will be equal to the rate of second reaction.

Therefore, differential rate laws can be written as follows,

$Rate of consumption of A = - \frac{d [A]}{dt} = k_{1} [A]$

$Overall, the rate of formation of B = \frac{d [B]}{dt} = k_{1} [A] - k_{2} [B]$

$Rate of formation of C = \frac{d [C]}{dt} = k_{2} [B]$

Solving the above differential rate laws,

${[A]}_{t} = [A_{0}] e^{- k_{1} t}$

${[B]}_{t} = \frac{k_{1} [A_{0}]}{k_{2} - k_{1}} (e^{- k_{1} t} - e^{- k_{2} t})$

Since the reaction is initially started with $A$ and the stoichiometric coefficients of $A$ , $B$ and $C$ are the same, so at any time in the reaction mixture, the sum of concentrations of $A$ , $B$ and $C$ will be always equal to $1 mol {dm}^{- 3}$ .

Hence,

${[A]}_{t} + {[B]}_{t} + {[C]}_{t} = [A_{0}] {[C]}_{t} = [A_{0}] - {[A]}_{t} - {[B]}_{t}$

Where,

$[A_{0}]$ is the initial concentration of $A$ .

Now, to calculate the composition of the mixture at time $t = 10 minutes$

Substitute $t = 10 mins = 0.17 {hr}^{- 1}$ {as rate constant have units of ${hr}^{- 1}$ }

and $[A_{0}]$ = $1.0 mol {dm}^{- 3}$

Concentration of $A$ ,

${[A]}_{t} = [A_{0}] e^{- k_{1} t} {[A]}_{t} = 1 \times e^{- 45 \times 0.17} mol {dm}^{- 3} {[A]}_{t} = 0.00048 mol {dm}^{- 3}$

Concentration of $B$ ,

${[B]}_{t} = \frac{k_{1} [A_{0}]}{(k_{2} - k_{1})} (e^{- k_{1} t} - e^{- k_{2} t}) {[B]}_{t} = \frac{45 \times 1}{(15 - 45)} (e^{- 45 \times 0.17} - e^{- 15 \times 0.17}) mol {dm}^{- 3} {[B]}_{t} = - \frac{45}{30} (0.00048 - 0.07808) mol {dm}^{- 3} {[B]}_{t} = 0.116 mol {dm}^{- 3}$

Now, the concentration of $C$ ,

${[C]}_{t} = [A_{0}] - {[A]}_{t} - {[B]}_{t} {[C]}_{t} = (1 - 0.00048 - 0.116) mol {dm}^{- 3} {[C]}_{t} = 0.883 mol {dm}^{- 3}$

Therefore, after a time interval of $10$ minutes, the composition (concentration) of the reacting system will be $[A] = 0.00048 mol {dm}^{- 3}$ , $[B] = 0.117 mol {dm}^{- 3}$ and $[C] = 0.883 mol {dm}^{- 3}$ .

Time in minute	$10$	$15$	$20$	$25$	$\infty$
Volume of $O_{2}$ given by $H_{2} O_{2}$	$6.30$	$8.95$	$11.40$	$13.5$	$35.75$

${[A]}_{t} = [A_{0}] e^{- k_{1} t} {[A]}_{t} = 1 \times e^{- 45 \times 0.17} mol {dm}^{- 3} {[A]}_{t} = 0.00048 mol {dm}^{- 3}$ (in a minute)	$0$	$75$	$119$	$259$	$\infty$
Titre value (in $mL$ )	$19.24$	$24.20$	$26.60$	$32.23$	$42.03$

Time (in minute)	$0$	$30$	$90$	$230$	$\infty$
Rotation (in degree)	$+ 46.75$	$+ 41.0$	$+ 30.75$	$+ 12.75$	$- 18.75$

Important Questions on Chemical Kinetics

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For a consecutive first-order reaction A⟶k1B⟶k2C the values of k1 and k2 are 45 hr-1 and 15 hr-1, respectively. If the reaction is carried out with pure A at a concentration of 1.0 mol dm-3, what will be the composition of the reacting system after a time interval of 10 minutes?

Important Questions on Chemical Kinetics

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