Question

4. Consider the reaction: 2 NO(g) + Br2 (g) 2 NOB (g) Kp = 31.8 at 302 K a) If the initial partial pressures are Pro = 108 torr, Per; = 126 torr, ProBr = 275 torr, find Qp and determine which direction the reaction proceeds in to reach equilibrium. 139.17 b) When a mixture of NO, Bra, and H, reaches equilibrium at 302 K, Pno=133 torr and Pers 151 torr. Find PNobr (in torr). [335 torr] 5. lodine gas and hydrogen gas are sealed in a flask and heated to 575 K. Before they react, the initial partial pressures of 12 and H, at 575 Kare: P =3.00 atm and PM, 2.00 atm. No other reactions take place. Find the equilibrium partial pressures of 12, H2, and HI at 575 K. 12(9) + H2(g) 2 HI(g) Kp = 92 at 575 K (P = 1.14 atm: Prz = 0.14 atm; Phi = 3.72 atm]

Answer 1

The direction of the equilibrium depends on Kc (equilibrium constant) and Qc reaction quotient). If K = Q, the reaction is said to be at equilibrium If K>Q, the amount of reactants is more than that of products. So the equilibrium will shift towards products. That is Forward reaction favours and the equilibrium will shift to right side. If K<Q, the amount of products is more than that of reactants. So the equilibrium will shift towards reactants. That is Backward reaction favours and the equilibrium will shift to the Left side. (a) Initial pressure of NO = 108 torr/ 760 = 0.142105 atm Initial pressure of Br2 = 126 torr/ 760 = 0.165789 atm Initial pressure of NOBr = 275 torr/ 760 = 0.361842 atm The reaction Quotient is 2 2.- Prope - (0.361842) Prox PB, = 39.1 (0.142105)° x(0.165789) The equilibrium constantis Kp = 31.8, which is less than Qp So, the equilibrium will shift towards reactants (left side)

The equilibrium pressures are Pro = 133/760 = 0.175 atm PBr2 = 151 /760 = 0.198684 atm KP = 31.8 Q: - photo = 31.8 PNo XP Biz ProB = 31.8x Pxo? x PB PROBE? = 31.8+(0.175)° x(0.198684) Probe? = 0.193494 PROBE = 0.439879 atm Pop = 0.439879 atm ~ 760 torr/atm = 334.3 toir