Question

3. Please use the rules for manipulating K above to answer the following questions. a) At 303 Kelvin, the equilibrium constant for the gas phase reaction of N, and Hą to form NH, is 2.8 x 10 What is the equilibrium constant for the decomposition of ammonia at the same temperature to produce hydrogen and nitrogen? Kreverse = Kforward 2.8x10-9 = 357142857.1 3.5x10 b) At 1000 Kelvin, the reaction Na(g) + 3H2(g) + 2NH3(g) has a kc value of 2.4 x 10° What is the value of Kc at 1000 Kelvin for the reaction 1/3 N (9) + H2(g) + 2/3 NH3(g)? c) Calculate the value of the equilibrium constant, K, for the reaction Q(g) + X(g) = M(9) using the following information: 2M(g) = 2(9) K = 6.2 x 104 Z(g) =2Q(9) + 2X(9) K2 = 5.6 x 10- Hint: How do you have to manipulate the above reactions so that they sum to the target reaction.

Answer 1

3. @ The equ expression written for a bocaetion written in the reverse direction is the reciprocal of the one for the forward meaction of k in the rate const of forward reaction k'n ann backward n then /kot/ l © N269+ 343(g)# 2N1blg) ke 2214810-3 at soook Keo 2.480-3 Mowo, 5 Maight HaCa) — % Ng(8) at doook ke', [Nirg]?/3 [Na] 5 [Hz] or, ke' (ke)%3 ke' - (2.481023)3 2 0.1338 © Qce) +(9) M(0) 2M()=7(9) koma for this Reaction ko [212 - 6.2x1004 2(g) = 22(g)+ax (g) mowe kaxky » TIPO Kg > [a]° [x12 os te maken TAJIET > Kyxky 2 0.338 1. [M72 2 25.6X10 LET - 2 162x10-4X5.6X102 0.00589 VkX1 2 169.77 Scanned with CamScanner

THE 5. ® CHA (9) + 120(g) + 3H2(g) + co(g) om constant ke Ke » [H219 [00] [eng] [HO] - keo (145) (0.126) 16 2846 | at 760°C (0-126)(0.242) 6 4(q) =23(g) at Conad vol“cud temp Kpop of two Pro Kp » (1.01) 14. 54 alm) 0.255 adequc 0.25 (1:32 -0.25 Dalm. WO) holato ICS: Scanding with ÇamScanner 35)

Answer 2

solution:- N2 (g) + 3H₂2NH3 g PNH .. PV=nRT P = 2 RT t 1 P=CRT WHz (RT)? (RT) * C H (RT)? CNH3 XC. (RT) gta 5 L (2 M2 gg + H 2 g) K2 2/3 NH3 (8) 22 QV oing on (Kr.) * = Ken =3) 2.4/1053 KE 0.18398

to 20 g + 2x8) K₂ = 5.6 x 16 K, = 6.2x104 o amg) = 20 K-k, 2 Mg) = 22,73+2x9) Divide the above equation by 2 Ma - Ag+xg) Kz_ki (5.6610²) (6. 2x104) k₂-ki 10-4 (560-612) 2 553.8x10-4 2. = 2.769 X 102 for agg + Yg) M (8) the equation becomes Hon becomes K, K2 2.769 XL = 36.114/1

k. PH₂ Pro PV=nRT - (CH2)" (Ceo) (RTS COTS (Cely (Cr2O) (B) Clip) = (CH2) (eo) X (RT)² (Certy) Cho R= 8.314 31 ki mole T = 760 +273 = 1033K (4.15)) (0.66) (9.814 x1033)? te (0.126) (0.242) [Ke= 4.635 x108 /

Ag) 220 1.320 Initial equilibrium 0.25 2.14 inget.og (axl.07 52.14 change of A is calculated by subtracting tega 0.25 from 1. 32 & change in B is calculated being by multiplying change change of of a a times 2 . PAIP to 1 = (2.14² 0.25