Question

I need help solving for number 2... I already have the answer so please teach me how to solve it. I want to understand.

the follwing dati dre for a system no P (AT) 0.000 0.120 0 180 0.110 0,350 0.350 ① . How does the rate of the fod rx n compare w tke rate of the reverse rxn affer 45 seconds? Aftee 902 Gets lower Pressure her pressere 45 → 90 :

Answer 1

K_eq = (P_B)^2 /(P_A) A Rightwards harpoon over leftwards harpoon^K 2B At 75 second equilibrium is reached K_eq = (0.350)^2 /0.325 = 0.377 At 45 second K = (0.280)^2 /0.360 = 0.217 At 90 second K = (0.350)^2 /(0.325) = K_eq = 0.377 as K is less than K_eq so the rate of farward reaction will be more after 45 seconds because K_eq = K_farward K_backward so, as A rightarrow 2B will increase pressure gets higher.