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

The formation in water of $d$ -potassium chromo-oxalate from its $l$ -form is a reversible reaction, which is first-order in both directions, the racemate being the equilibrium product. A polarimeter experiment at $22 ° C$ showed that after $560 s, 12 mole$ $%$ of the $l$ -isomer was converted to the $d$ -form. Find the rate constant for the forward and the reverse reactions.

Given $\ln (0.76) = - 0.2744$

Accepted Answer

Given, that a compound potassium chromo-oxalate has two enantiomeric forms $d$ and $l$ , which establish equilibrium with each other.

The reaction is started with $l -$ form, and at equilibrium, racemic mixture is formed, that means $d -$ form and $l -$ form are present in equal amount (moles) at equilibrium.

Therefore,

$l - form ⇌_{k_{b}}^{k_{f}} d - form; K_{eq}$

$K_{eq} = \frac{k_{f}}{k_{b}} = \frac{[d ‒ form]}{[l - form]} = 1$ (as the equilibrium product is a racemic mixture).

$\Rightarrow k_{f} = k_{b} . . . . . . . (1)$

Now, in the given reversible first-order reaction, after $560 s, 12 mole %$ of $l -$ form is converted into $d -$ form,

$l - form ⇌_{k_{b}}^{k_{f}} d - form t = 0 s n 0 t = 560 s n - 0.12 n 0.12 n,$

Where, $n = initial moles of l-form$ ,

$k_{f} and k_{b}$ are the rate constants of forward and backward reactions, respectively.

In the above given reversible reactions moles of product $(n_{t})$ at the time $t$ can be determined by the formula,

$n_{t} = \frac{k_{f} \times n}{k_{f} + k_{b}} \{1 - e^{- (k_{f} + k_{b}) t}\}$ .

From equation ( $1$ ), $k_{f} = k_{b}$

Let $k_{f} = k_{b} = k$

Now substituting the values, $n_{t} = 0.12 n mol$ and $k_{f} = k_{b} = k$ and $t = 560 s$ ,

$0.12 n = \frac{kn}{2 k} \{1 - e^{- 2 k \times 560}\} .$

Solving the above equation,

$0.12 = \frac{1 - e^{- 1120 k}}{2} 0.76 = e^{- 1120 k}$

Taking logarithm on both sides,

$ln (0.76) = - 1120 k k = 2.45 \times 10^{- 4} s^{- 1}$

Therefore, rate constants of forward as well backward reactions, $k_{f} = k_{b} = 2.45 x 10^{- 4} s^{- 1}$ .

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

$t$ (in a minute)	$0$	$75$	$119$	$259$	$\infty$
Titre value (in $mL$ )	$19.24$	$24.20$	$26.60$	$32.23$	$t = 560 s$

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

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