Three ideal gases A, B and D were taken in a vessel of constant volume in molar ratio of 1:2:3, resp β Chemical Equilibrium Chemistry Question
Question
Three ideal gases A, B and D were taken in a vessel of constant volume in molar ratio of 1:2:3, respectively, causing the reaction: A(g) + 2B(g) β C(g). After a long time, when the equilibrium is established, the total pressure was 5/6α΅Κ° of the initial total pressure. At this point, the volume was reduced to half and a catalyst was added starting the reaction: 2C(g) + D(g) β 2F(g). If at the second equilibrium, moles of A and C are equal and the ratio of equilibrium total pressure at second equilibrium to the equilibrium total pressure at the first equilibrium is 'x:100', then the value of 'x' is
π‘ Solution & Explanation
Let initial moles be A=1, B=2, D=3 (Total=6). Eq 1: A=1-y, B=2-2y, C=y. Total = 6-2y. Given P_eq1 / P_init = 5/6 => 6-2y = 5 => y = 0.5. Eq 1 moles: A=0.5, B=1, C=0.5, D=3. Total = 5. K_p1 for A + 2B β C requires calculating partial pressures in terms of P1. With V halved, reaction 2 starts, and reaction 1 shifts. Let rxn 1 shift forward by w, rxn 2 forward by z. Moles: A = 0.5-w, B = 1-2w, C = 0.5+w-2z, D = 3-z, F = 2z. Given n(A) = n(C) => 0.5-w = 0.5+w-2z => z=w. K_p1 = P_C / (P_A P_BΒ²) must remain constant. Since P_A = P_C, K_p1 = 1 / P_BΒ². This means P_B must equal its value at Eq 1 (which was 1/5 P1 = 0.2 P1). Total moles n_tot = 5 - 4w + z = 5 - 3w. P2 = (5-3w) Γ 2 Γ (P1/5). P_B at eq 2 = (1-2w)/n_tot Γ P2 = (1-2w) Γ 0.4 P1. Setting this to 0.2 P1: 0.4(1-2w) = 0.2 => 1-2w = 0.5 => w = 0.25. So z = 0.25. Total moles = 5 - 3(0.25) = 4.25. P2 = 4.25 Γ 2/5 P1 = 1.7 P1 = 170/100 P1. Thus x = 170. Therefore, correct answer is 0170.