Question

Ammonia gas is produced on a massive scale using the Haber-Bosch process. The reaction is: N2 (8) + 3H2 (8) À 2NH3 (8) AH = -92 kJ mol-1 (a) Write an expression for the equilibrium constant, Kp, for this reaction in terms of partial pressures. (b) Use the partial pressures given below to calculate the value of Kp at 700K, including its units. PM = 0.450 atmospheres P = 0.680 atmospheres Puu, = 0.152 atmospheres NIH (c) What would be the effects on the value of Kp, if the following changes were made to the system: (i) The partial pressure of nitrogen is increased by 0.100 atmosphere (ii) The temperature was increased to 980 K 5 Marks

Answer 1

N2(g) + 3H₂(g) + 2NH3 ig) term of partial Equilibrium constant pressure is written in as- Partial pressure of product produs essure of Kp = (partial pressur of reactant) Product Reactant ni m are Cotth trotren moves so) Now I feb (PNH)2 (pra) (62)3 kp = P NH3 = partial pressure of pressun of NH3 N2 PN2 = partial of pressure H₂ PM = partial

PN. 0.450 atm Ph68o at PNH (Pong ) 2 Kp = (Pm)" (PN) - (0.152 atm)? 6.4 soat) (o. 680 am) 3 x atma 152 X 152 x atm 680x uso X (0-6802 = 23104 - X(atm)-2 1414.94 auto 2 16.328 kp = ato c) W Not any effect on kp on There increasing presten of vessel. But, hear partial pressur of N2 h so by above formula Kp decreasest. фон onlу еатр- Ба крака -peratur of the ишон reaction. DH = -92 kJ ml + 3NH N2 + 3H₂ oh=-92 kJ molt This means reaction ŭ exothermic reaction les so on incresing formation of tempereration NH3- =) so less or deckesen decrease in kp.