Question

T. A reaction mixture at 350 T was found to have the following concentrations: [HI] 0.024 M; [H2] 0.025 M and [I2] 0.025 M H2(g) I2(g) 2 HI(g), K 57.0 a) Determine whether the mixture is at equilibrium or not b) If the mixture is not at equilibrium, tell the direction of the shift to reach equilibrium, and calculate the equilibrium concentrations of H2, I2 and HI 2. Consider the Haber Process for the synthesis of ammonia from hydrogen and nitrogen: 2 NH,(g) 3 H2(g)N2(g) Show that the reaction would shift to the right if the pressure is doubled 3. a) How would you drive the following reaction to the right to form more products CO2(g)H2 (g) 4H°=-40.3 k.J CO(g)H2O (g) Ans. b) One of the products is a 'high value substance'. Which one, what is it used for? and

Answer 1

Page No Solution (1) Given reaction: - en concentration O oa H2(g) + Iz (g) given concentration [HI]= 0.024 M [12] = 0.025 M = 2 HI(g) of each is [H2] = 0.025 M Kc = 57.0

Solutions) 1 Reaction quotient a concentrations of represents reactants varying and products. At equilibrium Kc = al so Q = [HI] ² [12] [12] = (01024) ? T0.025) (0.025) = 0.9216 Here Q LKC equilibriuto. so reaction is not at The mixture is hat at equilibriuto because + Kc

1b Here Q CKC Denominator (reactants) is larger than products (helmerator) Therefore reactants product. since more product is needed for Q = Kc So shift towards product for equilibrium & Direction forward gor gaining equilibrieem. Holg) + Tz (g) - 2HI 2HI 0.025 Change -n equilibrium 0.025-21 0.025 -n 0.025-2 0.024 20 0.024+2% Kc = [HI] [h] [I ] 57 = (0.024+22) 0.02 f0.025 0.025%) 57 = 0.024 +2n 0.025-2

0.024 +22 0.025-2 = 757 7.55 0.024 tan 000 25-a 0.024+2n = 7.55 0.025-2) 0.024 +22= 0. 1875 - 7.552 9,55% = OIBB75 -0.024 9.552 = 0.16475 n = 0.16475 9.55 n = 0.01725 (H₂) =0.025 -0.01725 ) M 6.75x10 3M TH2) = 7,758103 M ARS. (0.025-0001725 M (I2] = 7.75x10 39 ARS:

[HI] = 0,024 +(2X 0.01725) (HT) = 0.0625 M HI 0.05 B5 M Ans. Ans.