Question

The reaction AB(aq)→A(g)+B(g) is second order in AB and has a rate constant of 0.0157 M−1⋅s−1 at 25.0 ∘C. A reaction vessel initially contains 250.0 mL of 0.130 M AB which is allowed to react to form the gaseous product. The product is collected over water at 25.0 ∘C. How much time is required to produce 226.0 mL of the products at a barometric pressure of 718.3 mmHg . (The vapor pressure of water at this temperature is 23.8 mmHg.)

Answer 1

Ablaq) - A (g) + Blg) t 0 -71 moles of AB m a taken = 250 x0.13 = a Tood la= 0.0325 mm Total = 718.3 mm Hg = P + Pot PH 718.3 - PA+P = 718.3 – 23.8 = 694.5 mm 0.9138 aty Now, using PV=nRT, to get moles = 0.9138*226 = 2 xx 0.0821 x 298 1ooo 7 : 71 - 0.00422 moles of AB left = ax = 0.0325-0.00422 - 0.02828 A CAB) og = 0.0 2828 - = 0.1131 M 0.25

+ 0.0157x t uring and order a Integral eruation: to+kt ਹੈ । +6452 * > buenos 0052 ** - t = 73.21 Sec 0 1131 0.13 0.0157 xt 0.1131 0.13