## Answers

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.

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