Two chemicals a and b are combined to form a chemical

1 answer:

5 0

-dB/dt = kAB = k(2B)(B) = 2kB^2
-dB/B^2 = 2kdt
Integrating: 1/B - 1/(B_0) = 2kt
At t = 10, if 15 g of C have formed, this must have consumed 10 g A and 5 g B. The remaining mass of B is 45 g.
1/45 - 1/50 = 2(k)(10)
k = 1.11 x 10^-4
Then substituting this value of k with t = 40:
1/B - 1/50 = 2(1.11 x 10^-4)(40)
1/B - 1/50 = 0.008889
1/B = 0.028889
B = 34.62 g remaining
Therefore, 50 - 34.62 = 15.38 g of B have been consumed.
Doubling, 30.76 g of A have been consumed.
This means that 15.38 + 30.76 = 46.15 g of C have been formed.

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