Question

the equilibrium constant is given for two of the reactions below. Determine the value of the missing 23) - equilibrium constant A(g) +B(g) - AB(g) AB(g) + A(g) - A2B(g) 2 A(g) +B(g) - A2B(g) Ke=0.24 Ke = 3.8 Ke=? A) 0.63 B) 4.0 C) 16 D) 0.91 E) 3.6 24) 24) Determine the value of Kp for the following reaction, at 300K, if the equilibrium concentrations are as follows: (N2leq -1.5M, [H2leq = 1.1 M, (NH3 leq = 0.47 M. N2(8)+ 3 H2(8) - 2 NH3(g) A) 9.1 x 102 B) 3.5 x 10- 5 C ) 2.8 x 10-3 D) 1.8 x 10-4 E) 7.8 x 103 25) - 25) Determine the rate law and the value of k for the following reaction using the data provided. NO2(g) + O3(g) - NO3(g) + O2(g) [NO2l (M) 0.10 0.10 0.25 [O3l (M) 0.33 0.66 0.66 Initial Rate (M-1,-1) 1.42 2.84 7.10

Answer 1

24 23. Given scac are Alg) +813) = Ablg) K= 0.24 LAB YG) + Alg) = A2B(5) Rc = 3.8 A(3) +B19) = A2B (g) Ke=0.24+ 3.7 K4 s. Required Equilibrium constant Ke=4] Geiven that Nzig) + 3H2 (5) 22NH3) [N2] := 1.59 [1₂] = 1+1M [NH₂] = 0.47M. iKc = [NHg]² (0.47) - [N ] [H] (1.5)(1.13 20.11065 m -2 = 0.11065 Now . We know that 1 kp=Kq U T = 368 "P - An = (2 - (1+ 33. -0 . .: Rp = 0.11065(360 KX 0.08206 Lała 12 L = 1.826810-4 atm-2 .. Aus — D) 1.8810-47 35 Given sleach ei . No, (+ (3) - No G + (y Now, 8=K[No2]" [03]b Algiven clater. 1:42= K[0.10)*(0.336 -O 284 = K (0.10)° (0.66)6-

Dividing 0 80 ļ = 6) =) bal Similarly 284 = k10.107910.653b- 7.10 -K (0.7)9 (0.6.6J - © a:1 rate laws (8) = K [NO₂] a [og]b 1.42= klo.10) 10.33)' K = 43.04) 18 = K[NO₂] [og]]