Answer:
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
