a. Show that the difference A - B is constant in time. b. Use (a) to eliminate t

Question

a. Show that the difference A - B is constant in time.

b. Use (a) to eliminate the unkown B in the first equation of (1) and write down an initial value problem for A.

c. It was measured that 4 moles of reactant A remained after the reaction
                    had progressed for 1 second. What amounts of reactants A and B will be
                    left after 4 seconds.                 

Consider a chemical reaction of the form A + B = C, in which the rate of change of the two chemicals is described by dA / dt = -kAB; dB / dt = -kAB: (1) The positive constant k is a measure for the speed of the reaction. Assume that initially there are 5 moles of reactant A and 2 moles of reactant B. Show that the difference A - B is constant in time. b. Use (a) to eliminate the unknown B in the first equation of (1) and write down an initial value problem for A. It was measured that 4 moles of reactant A remained after the reaction had progressed for 1 second. What amounts of reactants A and B will be left after 4 seconds

Explanation / Answer

a)

dA/dt = dB/dt

-> dA = dB

-> int 1 dA = int 1 dB

-> A = B +c1

-> A - B = c1

where c1 is a constant.

at t = 0 : A(0) = 5 , B(0) = 2 -> c1 = 5-2 = 3

Therefore:

A-B = 3

b)

dA/dt = -kAB = -kA(A - 3)

A(0) = 5

c)

dA/dt = -kA(A-3)

1/A(A-3) dA = -k dt

int 1/3 * (1/(A-3) - 1/A) dA = int -k dt

1/3 (ln|A-3| - ln|A|) = -kt

t = 1 -> A = 4 -> 1/3 (ln(1) - ln(4)) = -k -> k = ln(4)/3 = (2/3)ln(2)

At t = 4:

1/3 (ln|A-3| - ln|A|) = -kt = -8/3 ln(2)

lnA/(A-3) = 8ln(2) = ln(128)

A/(A-3) = 128

1 - 3/(A-3) = 128

-3/(A-3) = 127 -> (A-3) = -3/127 -> A = 3 - 3/127 = 378/127 = 2.98

B = A - 3 = 0

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