Question

If, in Example 4, one molecule of the product C is formed from one molecule of the reactant A and one molecule of the reactant B, and the initial concentrations of A and B have a common value \(\left( A \right) = \left( B \right) = a{\rm{ moles/L}}\), then

\(\left( C \right) = \frac{{{a^2}kt}}{{\left( {akt + 1} \right)}}\)

where\(k\) is a constant.

(a) Find the rate of reaction at time \(t\).

(b) Show that if \(x = \left( C \right)\), then

\(\frac{{dx}}{{dt}} = k{\left( {a - x} \right)^2}\)

(c) What happens to the concentration as \(t \to \infty \)?

(d) What happens to the rate of reaction as \(t \to \infty \)?

(e) What do the results of parts (c) and (d) mean in practical terms?

Short answer

(a). \(\frac{{d\left( C \right)}}{{dt}} = \frac{{{a^2}k}}{{{{\left( {akt + 1} \right)}^2}}}\)

(b). \(\frac{{dx}}{{dt}} = k{\left( {a - x} \right)^2}\)

(c). Concentration becomes \(a{\rm{ moles/L}}\)

(d). Rate of reaction will be zero.

(e). The reaction would virtually stops.

Step-by-step solution

Concentration of a Solution

The Concentration of a solution can be defined as the presence of amount of solute in the whole solution, where the amount of solute is taken in moles and the whole solution is taken in litres.

Evaluating the given function using differentiation:

(a)

Thegiven function is:

\(\left( C \right) = \frac{{{a^2}kt}}{{\left( {akt + 1} \right)}}\)

Now, differentiating with respect to t:

\(\begin{aligned}\frac{{d\left( C \right)}}{{dt}} &= \frac{d}{{dt}}\left( {\frac{{{a^2}kt}}{{\left( {akt + 1} \right)}}} \right)\\ &= \frac{{\left( {akt + 1} \right)\frac{d}{{dt}}\left( {{a^2}kt} \right) - \left( {{a^2}kt} \right)\frac{d}{{dt}}\left( {akt + 1} \right)}}{{{{\left( {akt + 1} \right)}^2}}}\\ &= \frac{{{a^2}k\left( {akt + 1 - akt} \right)}}{{{{\left( {akt + 1} \right)}^2}}}\\ &= \frac{{{a^2}k}}{{{{\left( {akt + 1} \right)}^2}}}\end{aligned}\)

Hence, this is the required rate of reaction.

Examining the given function of reaction: 

(b)

Since, \(x = \left( C \right)\)

Therefore, we have:

\(\begin{aligned}a - x = a - \left( C \right)\\ &= a - \frac{{{a^2}kt}}{{\left( {akt + 1} \right)}}\\ &= \frac{a}{{\left( {akt + 1} \right)}}\end{aligned}\)

Then,

\(\begin{aligned}k{\left( {a - x} \right)^2} &= k{\left( {\frac{a}{{akt + 1}}} \right)^2}\\ &= \frac{{{a^2}k}}{{{{\left( {akt + 1} \right)}^2}}}\\ &= \frac{{d\left( C \right)}}{{dt}}\\ &= \frac{{dx}}{{dt}}\end{aligned}\)

Hence proved, \(\frac{{dx}}{{dt}} = k{\left( {a - x} \right)^2}\).

Evaluating for concentration:

(c)

Now as \(t \to \infty \) we have:

\(\begin{aligned}\mathop {\lim }\limits_{t \to \infty } \left( C \right) &= \mathop {\lim }\limits_{t \to \infty } \left( {\frac{{{a^2}kt}}{{\left( {akt + 1} \right)}}} \right)\\ &= \mathop {\lim }\limits_{t \to \infty } \left( {\frac{{\frac{{{a^2}kt}}{t}}}{{\frac{{\left( {akt + 1} \right)}}{t}}}} \right)\\ &= \mathop {\lim }\limits_{t \to \infty } \left( {\frac{{{a^2}k}}{{\left( {ak + \frac{1}{t}} \right)}}} \right)\\ &= \frac{{{a^2}k}}{{ak}}\\ &= a\end{aligned}\)

Hence, concentration becomes \(a{\rm{ moles/L}}\) as \(t \to \infty \).

Evaluating for rate of reaction: 

(d)

As \(t \to \infty \) we have:

\(\begin{aligned}\mathop {\lim }\limits_{t \to \infty } \frac{{d\left( C \right)}}{{dt}} &= \mathop {\lim }\limits_{t \to \infty } \left( {\frac{{{a^2}k}}{{{{\left( {akt + 1} \right)}^2}}}} \right)\\ &= 0\end{aligned}\)

Hence, the rate of reaction becomes zero as\(t \to \infty \).

Explaining practically regarding part (c) and (d):

(e)

As \(t \to \infty \), approximately all the reactants A and B have been converted into the product C. As a result, no further reaction takes place as t

increases.

Hence, this is the required answer.