Question

Solve this question using the methot used in below example 8.1

DO NOT USE ANY OTHER METHOD, ONLY USE THE ONE THAT HAS BEEN USED IN EXAMPLE 8.1
And please include all the steps

thanks in advance!

2. For the reaction given below qA(O) 18,0)p(0)-0,k 10,k2 4 Solve for qa k2 ap EXAMPLE PROBLEM 8.1 Consider the reaction given in Eq (8.14), with qa(0) -10,4(0) -15, and go(0)-0, K -2 From Eq. (8.14), it is clear that ip---4 and after integrating, we have To solve for ge, we eliminate qu and ga using Eq. (8.15) by substituting qA ga(0)+4P(O)-q and K-1-3. Solve for qp Solution (8.15) and qB = qB(0) + qr(0)-qv into Eq. (8.14), giving 4p K1QAqB-K. iqp # K1(qv(0) + qa@)-9P)(qr(0) + qr(0)-qr) _ K-14P (8.16) Since gp(0) - 0, we have = 2(10-9p)(15-4»)-39P (8.17) 53 and after rearranging terms dap 呢ー芝9p + 150 Once again, partial fraction expansion is used to rewrite Eq (8.18) as 0.098o(da dap (8.18) (4p-18.309 , _ 8.2)=2.0 p- 8.2) (8.19) (qr 18.3) (qp-8.2) Contimued where B1 = | (9p-18.3) =0.0989 18.3-14,-82)|| P-82カlar- t(9p-8.2 -18.3 and IGp-82) (ap-183)(邪ー8.2)1-2-19p-18.3) =-0.0989 Integrating Eq. (8.19) and rearranging yields In(22.9p-82) (@pー18.3) =ー20.2t and after solving for qp, we have 82(1 e-20.2 @p = (8.20) The solution for the reactants are found from gA 44(0)+ap(0) -9p and 1.8059 +3.7e-202 and 1-0.45e02 u(t) 6.8059 + 1.4798e-22 4a = (0.51 1-0.4476e: 202tー)11(1) These results are shown in Figure 8.2. Note that conservation of mass is still maintained, since qr contains both ga and qa, so gA +q +2xqe 25.

Answer 1

46 SOLUTION kahon ftr , we nud to 11.ouhahe qf and ใ b Novw 4 ) 12S t 186 Ro