Assignment 5: Problem 6 PreviouS Problem List Next (1 point) Two chemicals A and

Question

Assignment 5: Problem 6 PreviouS Problem List Next (1 point) Two chemicals A and B are combined to form a chemical C. The rate of the reaction is proportional to the product of the instantaneous amounts of A and B not converted to chemical C. Initially there are 77 grams of A and 93 grams of B, and for each gram of B, 1.8 grams of A is used. It has been observed that 42.5 grams of C is formed in 15 minutes. How much is formed in 20 minutes? What is the limiting amount of C after a long time? grams of C are formed in 20 minutes grams is the limiting amount of C after a long time

Explanation / Answer

ASSIGNMENT 5:PROBLEM 6

ans: 51.42 g of C are formed in 20 minutes

119.78g is the limiting amount f C after long time  

SOL: Let X(t) be the function which gives the instantaneous value of C

For each gram of B , 1,8 g of A Is used =>there are total 2.8 parts(OUT OF WHICH 1 PART CONSTITUTES B ,1.8 PARTS CONSTITUTES A)

The instantaneous values of A and B are

77-(1.8/2.8)X , 93-(X/2.8) respectively

as the rate of reaction is propotional to instantaneous values of A and B

dX/dt =k(77-(1.8/2.8)X)(93-(X/2.8))

Solving the differential equation

dX/dt =k(1.8/(2.8*2.8))(119.78-X)(260.4-X)

dX/(119.78-X)(260.4-X) =0.23kdt

splittng into partial fractions

(dX/(119.78-X))-(dX/(260.4-X))=0.23kdt/0.007

Applying intergration on both sides

ln|260.4-X|-ln|119.78-X|=32.80kt+c

260.4-X(t)/119.78-X(t) =c' e^(32.80kt) -> consider this as equation 1

we know that from the given problem

X(0)=0 X(15)=42.5 g

substituting X(0)=0 ,we get c'=2.17

now by

substituting c' and x(15)=42.5g

we get value of k as 0.000531

260.4-X(t)/119.78-X(t) =2.17e^(0.017t)

rearranging the terms to get X(t)

X(t) = (259.92-260.4 *e^(-0.017t))/(2.17-e^(-0.017t))

the amount of C after 20 minutes is obtained by substituting t=20 in the above equation

we get X(20)= 51.42 gms

limiting value of C after long time is obtained by substituting t=infinity in equation of X(t)

as t->infinite the value of X(t) become 119.78g

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