Solutions and colligative properties and chemical kinetics combined ques.

[sirisha kotik]

a mixture of two miscible liquids a nd b has 10 moles of a and 12 moles of b. pure vapour pressure of a is 300mm Hg and tht of b is 500mm of hg. as soon as b is added to solution A starts polymerising into an insoluble solid. after 100 minutes 0.525 moles of a solute C is added whch stops polymerisation completely.the final vapour pressure of solution is 400 mm hg. calculate rate constant of the polymerisation reaction. its follows first order kinetics.

Homework Equations

k=1/t ln(c1/c2) p= p1x X1 + p2x X2

The Attempt at a Solution

ok first calculated the total v.p of initial solution. mole frac. of A is 5/11 and tht of B is 6/11 so total v.p comes to be 300x5/11+500x6/11 whch is equal to 4500/11. after tht by using pressure is proportional to moles calculated final moles of last solution which is coming abt 21.51 and initial moles are 22. but ,main problem is i m not gettng how to calculate moles of A that have polymerised. please help.

 

[nate808]

To calculate the rate constant of the polymerization reaction, we first need to determine the moles of A that have polymerized. We can do this by using the ideal gas law, which states that the pressure of a gas is directly proportional to the moles of the gas present, assuming constant temperature and volume. In this case, we know the initial pressure of A (300 mm Hg), the initial moles of A (10), and the final pressure of A (400 mm Hg). Using the ideal gas law, we can calculate the final moles of A as follows:

P1V1 = P2V2 (assuming constant temperature and volume)

(300 mm Hg)(V1) = (400 mm Hg)(22 moles)

V1 = (400 mm Hg)(22 moles)/(300 mm Hg) = 29.33 moles

Therefore, the moles of A that have polymerized are 10 - 29.33 = -19.33 moles. This negative value indicates that A has been consumed in the reaction.

Now, we can use the given information about the solute C to calculate the rate constant of the reaction. Since the reaction follows first-order kinetics, we can use the integrated rate law for a first-order reaction:

ln[C]t/[C]0 = -kt

where [C]t is the concentration of C at time t, [C]0 is the initial concentration of C, k is the rate constant, and t is time.

We know that after 100 minutes, 0.525 moles of C have been added and the reaction has completely stopped polymerization. Therefore, [C]t = 0.525 moles and [C]0 = 0 moles. Plugging these values into the integrated rate law, we can solve for k:

ln(0.525/0) = -k(100 minutes)

k = -(ln(0.525/0))/(100 minutes) = 0.0053 minutes^-1

Therefore, the rate constant of the polymerization reaction is 0.0053 minutes^-1.