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 10 grams of C is formed in 5 minutes. How much C is formed in 5 minutes. How much C is formed in 20 minutes

At 5 minutes, 10g is formed

At 20 minutes, 29.321g is formed

Step-by-step explanation:

Let X(t) represent the number of grams of Compound C present at time (t).

From the question, for each gram of B, 2 grams of A are used. Thus;for X grams of C, we have;

(2/3)X grams of A and (1/3)X grams of B.

Hence, the amounts of A and B remaining at any given time is;

40 - (2/3)X grams of A and 50 - (1/3)X grams of B. Now, we know that the rate at which compound C is formed satisfies;

dx/dt ∝ (40 - (2/3)X)(50-(1/3)X) which gives;

dx/dt = k (120 - 2X)(150 - X)

dx/[(120 - 2X)(150 - X)] = kdt

∫dx/[(120 - 2X)(150 - X)] = ∫kdt

Integrating, we have,

In[(150-X)/(120-2X)] = 180kt + C

Simplifying further,

[(150-X)/(120-2X)] = Ce^(180kt)

By using,

X(0) = 0, we get;

[(150-0)/(120-2(0))] = Ce^(0)

C = 150/120 = 5/4

Now, pligging it intonthe equation to get ;

[(150-X)/(120-2X)] = (5/4)e^(180kt)

To find k, from the question, X(5) = 10.thus;

[(150-10)/(120-2(5))] = (5/4)e^(180k x 5)

140/100 = (5/4)e^(180k x 5)

1.4/1.25 = e^(900k)

1.12 = e^(900k)

In 1.12 = 900k

900k = 0.11333

k = 0.11333/900 = 1.259 x 10^(-4)

So,for x(20), and plugging in the value of k, we have;

[(150-X)/(120-2X)] = (5/4)e^(180 x 1.259 x 10^(-4) x 20)

[(150-X)/(120-2X)] = 1.9668

150 - X = 1.9668(120-2X)

150 - X = 236.016 - 3.9336X

3.9336X - X = 236.016 - 150

2.9336X = 86.016

X = 86.016/2.9336 = 29.321g

for b its (t) and c its (d)

step-by-step explanation: