Question

1. Two chemicals A and B react to form another chemical C. The rate at which is formed varies as the product of the instantaneous amounts of A and B present. For each gram of B the reaction requires 2g of A to produce C. 10g of A and 20g of B are present initially; there is no C present initially, and 6g of Care formed in 20 min. (a) Show that æ(t), the amount (g) of C present at time t (hours), is determined by the differential equation z' = k(15 - x)(60 – 2), where k is the rate parameter. (b) Solve the equation to obtain r(t), using the information provided above. (c) What is the limiting value for the amount of C formed?

Answer 1

( Reaction) set PATB • 3C let amount of a at any time uth then RA 3 Amount eac (9) of A at any time 2 x Amount of B at any time in amount of A AA ( 10 - 2x 3 (15-x) Change Change in Amount of Bo Chemical Rate Recation AB 20-25 z ¢(60-x) is : dc dt L. ΔΑ: ΔΒ = 4k (15-) 660-1) 9 du dt -k (15-x) (0o-su) E qA x cod=09 acao) 63 b dx_k(15-2 (60-1) dt Systems of kit (15-se] (oo-n] or 45 15-

+ ln (154) 45 tl In (Gomes aktt Constant 0 In bone 15-1 kt - Constant 45 In 160-x 45 kt - es construit 15-1 When Hag to In ( 6 Je constant 1nr6o J = 45kt Go an + 15-a IS óta Inl Go Goor 151 J = 45K us Go afskt -k x 15 60 e bot Go 45kt Ison 15-n 15 iskt to-n ze oo - Goxetskt 15 =) 6o Cl-efskt) a 1-toe toket 15 xt) = -60 61-e45kt) - 4 e 4skt

But Ka.at 여 4sceming actJ = 60 (-e letet . lokty Hets= Go (lae elokt at roo tz domin = I hours (1-4 2 3 0 6 3 z Go olej 0 10 . (l-e (1-4 elok Ka 3 In 10 () xect) = 60 (1 - et ne sy (1-4 eth (es) ? «GtJz 60 (1-BIJE (1-4 Cent) OR

(C) Limiting amount xcts Lim t to t s hranit as Lem N- t-poo se Go O 27 (1-4(27jt) Lim 6o (( Jt - 1) (-4) Limit too Go Co- Co-4) 15 g
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