Question

please solve all the parts, so I can check the answers I already have. Thanks!

Consider a reaction which proceeds only the forward direction A→ 3B S and has the following rate law d[A] _ -ky[A]Ž a. Derive an expression for [A] as a function of time Jb. You perform some experiments using this reaction at T = 1000.0 °C. If the initial concentrations (t = 0 s.) are [A]. = 0.133 [Bl. = 0.025 and at t = 87.0 s. [B] = 0.101 Calculate the concentration of [A] at t = 55.0 s. Vc. Assuming the same conditions as part b, at what time will [B] = 0.129 d. You perform similar experiments at 1200.0 °C and find that if [A], = 0.157 at t = 83.0 s. the concentration is [A] = 0.094 Using the Arrhenius equation determine the activation energy and the frequency factor of this reaction?

Answer 1

d from conation we have. ${chin-Cas } = *** ${cology for sayin} = *,* 23 * kr (af 1200°c) = 44.444154 Nor, we have krat 1000°c) = 17.284104 Kr (at 1200°c) - 44.46 xlony we know. kr=A. EN/RT where é where fa activation energy. - Ea/RT Cat 1000) (km, (at 1000c) (kr) (at 1200c) a talRT (1200%) in (kamkan) - Ed ( 1 ) as to 1975) on, fo.9445 = 7 Ea x 10.677105 8.314 Ea = 0.73594100 J mor - 0.73596 182 KJ mo! 73.59 & J mot!

he given reaction in ABB Now, rate law - d[A] de = kg EA73/ dLAS - Krdt [A32 Integrating both Sidle, we get to - dla [A] = kr (do on [A] = kp[tit m, ti capire choix} = Kit.... slope=2kr Given [A] = 0.133 [B]g7.05 0.101 [B] = 0.025 concentration B increased by (0.101-0.025) - 0.076 e Concentration of A decreased by. 0.076 [A] 07.05 (0:133-0.025) 20.108 = 0.025 from equation we get 09 ${idterin - 6-1203,4} = k1X270 or k= 17.28x1004 unit

from equation ③ we have, A = 2 krt & A - (2x 17-28 X10*x55 ) + fo 125mm 0.190 +2.74 - 2.93 . [1] s.Os = 1 = 0.341 on [A]ssos = (0.341% = 0.116 Concentration of CA) at 55.0s is 0.116 ③ Given i. [B] = 0.129 [AB = 0.125 = 0.043 (as 3 moles of B is. from one more i = 0.043 L formed Concentration AS A of A at time t= 0.133-0.0 from equation ③ we get 0.09 * ${co-ogh (1973's} = 17.28x10*x+ t = 171.1$. required time = 171.1 s.