Question

6. Use the reaction:NH N03 () 2H20 (q) + N20 ) to answer the following questions. a) (5 pts) Set up the Kc equilibrium expression for the above reaction. b) (5 pts) If Kc for the reaction is 584.7 at 523 K, what is Kp for the same reaction? C) (5 pts) If at equilibrium, there is 106 g of NH4NO3, and the concentration of H20 is 11.2 M, what would the concentration of N20 be? d) (5 pts) If the reaction is exothermic, describe how increasing the Temperature, Pressure, amount of NH4NO3 and adding a catalyst will affect the amount of the product, N2O? E this xtra credit: what does symbol "t mean?

Answer 1

6) NH4Nz Ž N2O + 2H₂O a) | Kc = [+0]² TN20] solids do not have concentration b) Ang = np-RR = 3-0 73 ke=584.7 ; Kp = KART) Ang = 584. 7 (00821 X 523)

kp=46287916.39 kp= 4.626x1071 c) RHINO = 106g ANANO3 = 106g Bogmort molar mass = 1.325mol (tholegin = 11.2M NHy A103 = NO +24,0 to: n - - egin: n-n x 22 [ro legim = 11.2M = 22 5.6M - 11.2M 2 - Inolegin = 5.6M I d) NH4NO3 = 2440+N,0; 6H--ve

D. On increasing temperature equilibrium will try to nullify the effect. so equilibrium will shift in endothermic reaction, that is reactant side and concentration of product will DECREASE .

On increasing pressure, equilibrium will try to shift to nullify that affect. it will shift in the direction where number of moles are less and hence less pressure is present. so equilibrium will shift in reactant direction because pressure is less there due to less number of moles. Hence, product concentration will again DECREASE.

On increasing number of moles of ammonium nitrate, product concentration increase because it will drive equilibrium towards products side.

Addition of catalyst has no effect on equilibrium constant value but it help in achieving the equilibrium faster. it accelerates the rate of reaction. No effect on concentration of either reactant or product.