Two chemicals, A and B, are combined, forming chemical C. The rate of the reacti

Question

Two chemicals, A and B, are combined, forming chemical C. The rate of the reaction is jointly proportional to the amounts of A and B not yet converted to C. Initially, there are 200 grams of A and 300 grams of B and, during the reaction, for each of gram of A used up in the conversion, there are three grams of B used up. An experiment shows that 75 grams of C are produced in the first ten minutes, the amount of chemical C, X(t), produced by timed is: Select the correct answer and please explain. Thanks!

a) X = 800(1 - e-400kt)/(2 - e-400kt) , where k = ln(29/26)/400

b) X = 800(1 - e-800kt)/(4 - e-800kt) , where k = ln(29/20)/8000

c) X = 400(1 - e-800kt)/(4 - e-800kt) , where k = ln(29/20)/8000

d) X = 400(1 - e-400kt)/(4 - e-400kt) , where k = ln(29/26)/4000

e) X = 600(1 - e-800kt)/(4 - e-800kt) , where k = ln(29/20)/8000

Explanation / Answer

A + 3B ----> C

for x grams of C , 200-x/4 of A , 300 -3x/4 of B is used up

dx/dt = K ( 200 - x/4) (300- 3x/4)

every time we make one gm of C we use up three gms of B

now

dx / ( 200 - x/4) (100- x/4) = 3k dt

integrating we get

ln {( x-800)/(x-400) } = 300*k*t/16 + ln P

whre P is integration constant

(x-800) /(x-400) = P*e^(300kt/16)

at t=0, x=0, start of reaction

so P=2

now

(x-800) /(x-400) = 2*e^(300kt/16)

x-800 = 2*x*e^(300kt/16) - 800*e^(300kt/16)

x ( 1-2*e^(300kt/16)) = 800 (1-e^(300kt/16))

x = 800 ( 1- e^(-300kt/16)) / ( 2- e^(-300kt/16) )

at t=10, x= 75

substitute in (x-800) /(x-400) = 2*e^(300kt/16)

we get   1/2 * 725/325 = e^(3000k/16)

3000k/16 = ln(29/26)

k= 16*ln(29/26)/3000

put in x = 800 ( 1- e^(-300kt/16)) / ( 2- e^(-300kt/16) ) we get

x = 800 ( 1- e^(-ln(29/26)*t/10)) / ( 2- e^(-ln(29/26)*t/10) )

x = 800 ( 1- e^(-ln(29/26)*t/10)) / ( 2- e^(-ln(29/26)*t/10) )

a) X = 800(1 - e-400kt)/(2 - e-400kt) , where k = ln(29/26)/4000

answer is A

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