Two chemicals A and B are combined to form a chemical C. The rate, or velocity,

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 100 grams of A and 50 grams of B, and for each gram of B, 2 grams of A are used. It is observed that 10 grams of C is formed in 5 minutes.

a) How much is formed in 20 minutes?

b) What is the limiting amount of C after a long time?

c) How much of chemicals A and B remains after a long time?

d) At what time is the chemical C half-formed?

Points will be awarded to the person who has the right answer with all work shown.

Explanation / Answer

This problem is made easy by having the original ratio of A and B equal to the rate at which A and B are used. This allows us to eliminate one variable. Then dC/dt = k B^2

Yet, as delta C = delta B + delta A = delta B + 2 delta B = 3 delta B, we may rewrite this as dB/dt = -k/3 B^2

In 10 minutes, if 10 g of C is formed, 10/3 g of B is used up. We may replace k/3 with c

Then, dB/B^2 = c dt, or

-1/B = ct + y

B = -1/(ct+y)

At t = 0, we have B = 50, so -1/50 = y

Then, at t = 10, B = 50 - 10/3 = 140/3, and -1/(140/3) = c*10 - 1/50

-3/140 + 1/50 = 10 c

-1/700 = 10c

c = -1/7000

Then, B = -1/(ct+y) = 1/(1/7000t + 1/50), or B = 7000/(t + 140)

Then, in 20 minutes, B = 7000/(t + 140) = 7000/160 = 175/4

Then, C = 3 * (50 - 175/4) = 75/4 = 18.75

b) The limiting amount is simply at B = 0.  C = 3 * (50 - 0) = 3 * 50 = 150.

c) The limiting amount of A and B is 0.

d) The chemical C is half-formed when half ob B is used up, so B = 25

Then, 25 = 7000/(t+140)

Cross-multiplying, 25t + 3500 = 7000

25t = 3500

t = 140

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