Question

Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of A and B not converted to chemical C. Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams of C is formed in 7 minutes. How much is formed in 14 minutes? What is the limiting amount of C after a long time? How much of chemicals A and B remains after a long time?

Answer 1

Answer:

25.35 grams of C  is formed in 14 minutes

after a long time , the limiting amount of C = 60g ,

A = 0 gram

and  B = 30 grams;           will remain.

Step-by-step explanation:

From the information given;

Let consider x(t) to represent the number of grams of compound C present at time (t)

It is  obvious that x(0) = 0 and x(5) = 10 g;

And for x gram of C;

$\dfrac{2}{3}x$   grams of A is used ;

also $\dfrac{1}{3} x$   grams of B is used

Similarly; The amounts of A and B remaining at time (t) are;

$40 - \dfrac{2}{3}x$   and $50 - \dfrac{1}{3}x$

Therefore ; rate of formation of compound C can be said to be illustrated as ;

$\dfrac{dx}{dt }\propto (40 - \dfrac{2}{3}x)(50-\dfrac{1}{3}x)$

=$k \dfrac{2}{3}( 60-x) \dfrac{1}{3}(150-x)$

where;

k = proportionality constant.

= $\dfrac{2}{9}k (60-x)(150-x)$

By applying the  separation of variable;

$\dfrac{1}{(60-x)(150-x)}dx= \dfrac{2}{9}k dt$

Solving by applying partial fraction method; we have:

$\{ \dfrac{1}{90(60-x)} - \dfrac{1}{90(150-x)} \}dx = \dfrac{2}{9}kdt$

$\dfrac{1}{90}(\dfrac{1}{x-150}-\dfrac{1}{x-60})dx =\dfrac{2}{9}kdt$

Taking the integral of both sides ; we have:

$\dfrac{1}{90}\int\limits(\dfrac{1}{x-150}- \dfrac{1}{x-60})dx= \dfrac{2}{9}\int\limits kdx$

$\dfrac{1}{90}(In(x-150)-In(x-60)) = \dfrac{2}{9}kt+C$

$\dfrac{1}{90}(In(\dfrac{x-150}{x-60})) = \dfrac{2}{9}kt+C$

$In( \dfrac{x-150}{x-60})= 20 kt + C_1 \ \ \ \ \ where \ \ C_1 = 90 C$

$\dfrac{x-150}{x-60}= Pe ^{20 kt} \ \ \ \ \ where \ \ P= e^{C_1}$

Applying the initial condition x(0) =0  to determine the value of P

Replace x= 0 and t =0 in the above equation.

$\dfrac{0-150}{0-60}= Pe ^{0}$

$\dfrac{5}{2}=P$

Thus;

$\dfrac{x-150}{x-60}=Pe^{20kt} \\ \\ \\ \dfrac{x-150}{x-60}=\dfrac{5}{2}e^{20kt} \\ \\ \\ 2x -300 =5e^{20kt}(x-60)$

$2x - 300 = 5xe^{20kt} - 300 e^{20kt} \\ \\ 5xe^{20kt} -2x = 300 e^{20kt} -300 \\ \\ x(5e^{20kt} -2) = 300 e^{20kt} -300 \\ \\ x= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

Thus;

$x(t)= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

Applying the initial condition for x(7) = 15 , to find the value of k

Replace t = 7 into $x(t)= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

$x(7)= \dfrac{300 e^{20k(7)}-300}{5e^{20k(7)}-2}$

$15= \dfrac{300 e^{140k}-300}{5e^{140k}-2}$

$75e^{140k}-30 ={300 e^{140k}-300}$

$225e^{140k}=270$

$e^{140k}=\dfrac{270}{225}$

$e^{140k}=\dfrac{6}{5}$

$140 k = In (\dfrac{6}{5})$

$k = \dfrac{1}{140}In (\dfrac{6}{5})$

k = 0.0013

Thus;

$x(t)= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

$x(t)= \dfrac{300 e^{20(0.0013)t}-300}{5e^{20(0.0013)t}-2}$

$x(t)= \dfrac{300 e^{(0.026)t}-300}{5e^{(0.026)t}-2}$

The amount of C formed in 14 minutes is ;

$x(14)= \dfrac{300 e^{(0.026)14}-300}{5e^{(0.026)14}-2}$

x(14) = 25.35 grams

Thus 25.35 grams of C  is formed in 14 minutes

NOW; The limiting amount of C after a long time is:

$\lim_{t \to \infty} = \lim_{t \to \infty} (\dfrac{300 e^{(0.026)t}-300}{5e^{(0.026)t}-2})$

$\lim_{t \to \infty} (\dfrac{300- 300 e^{(0.026)t}}{2-5e^{(0.026)t}})$

As; $\lim_{t \to \infty} e^{-20kt} = 0$

⇒ $\dfrac{300}{5}$

= 60 grams

Therefore  as t → $\infty$;   x = 60

and the amount of A that remain = $40 - \dfrac{2}{3}x$

=$40 - \dfrac{2}{3}(60)$

= 40 -40

=0 grams

The amount of B that remains = $50 - \dfrac{1}{3}x$

= $50 - \dfrac{1}{3}(60)$

= 50 - 20

= 30 grams

Hence; after a long time ; the limiting amount of C = 60g , and 0 g of A , and 30 grams of B will remain.

I Hope That Helps You Alot!.