decomposition of ammonium carbonae: 20Nlcoo)2 altherrnie (AF-o) reaction. In whi

Question

decomposition of ammonium carbonae: 20Nlcoo)2 altherrnie (AF-o) reaction. In which direction will the position endothermic 2N11,(x) + Code) , Hodg) is an h of the following changes (shift Ict, riaht, or no shanus) m will the position of the equitibrium be a. NH-(e) is removed: H:08) is added: . Part of (NH) COs(o) is removed d. The volume of the container is halved . The temperature is increased (the reaction is endothermic)y 2NH() At227 °C, Ke-1.00×102 for the reaction: Calculate values of K. for the following reactions at this temperature 2N2(g) + 6Hdg) 4NH3(g) 5. Nag) + 3Htg) a. c. Calculate Kp (at 227 C) for the reaction: 2NH()N()+3H:() R 0.08206 L-atm/mol-K]

Explanation / Answer

Ans 5

Part a

Multiply by 2 in the original equation, gives

2N2 + 6H2 = 4NH3

Kc' = Kc2 = (1*10^-2)2

Kc' = 10^-4

Part b

Reverse and multiply by 1/2 in the original equation

Kc'' = (1/Kc)1/2 = (1/10^-2)0.5

Kc" = 10

Part C

Kc for the 2NH3 = N2 + 3H2

Kc = (1/10^-2) = 100

Kp = Kc (RT)^n

Kp = 100 (0.08206*500)^(4-2)

= 168346.09 = 1.683 x 10^5

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