Question

6. In this experiment 2.00 mols of NOCI are placed in a 1.00L equilibrium is achieved, nitrogen monoxide NO and chlorine gas equilibrium concentration of NO is 0.66 M. Calculate Kc for this placed in a 1.00 L flask @ 298 K. After ad chlorine gas Cl, form. The alculate Kc for this equilibrium reaction: ? 2 NOCI (g) = 2 NO(g) + Cl2 (8)

Answer 1

6.

2 NOCl (g) $\rightleftharpoons$ 2 NO (g) + Cl₂ (g)

Now, [NOCl] = (2.00/1.00) = 2.00 M

ICE table is

Concenteation	[NOCl] , M	[NO], M	[Cl2] , M
Initial	2.00	0	0
Change	(0.66 - 2.00) = - 1.34	+1.34	+(1.34/2) = 0.67
Equilibrium	0.66	1.34	0.67

Now,

Kc =
_{[NO] [C12] [N OC112}

Or, Kc =
_{(1.34)2 x (0.67) (0.66)2}

Or, Kc = 2.76

3.

2 HI (g) $\rightleftharpoons$ H₂ (g) + I₂ (g)

K_p =
_{PH, X P12 P?}

K_p =
_{0.0960 x 0.0960 (0.708)2}

K_p = 0.018