Equilibrium Chemistry (Can Someone Solve This)

Question

2H₂Cl_(g) <--> 2H_2(g) + Cl_2(g)

A 4.30 g sample of H₂Cl_(g) is introduced into an evacuated rigid 1.50 L container. Upon heating to 423 K, hydrogen chloride gas decomposes, and 0.0130 moL of Cl_2(g) is present at equilibrium.

a. Write the expression for the equilibrium constant, Kc, for the decomposition reaction represented above.

b. Calculate the equilibrium concentration in molarity of the following gases in the container at 423 K

(i) H_2(g)

(ii) H₂Cl_g)

c. Calculate the value of the equilibrium constant, K_c, for the decomposition reaction at 423 K

d. Calculate the equilibrium constant in terms of K_p, for the reaction at 423 K

e. Calculate the partial pressure of Cl_2(g) in the container at equilibrium at 423 K

Answer 1

Initial moles HCl = 4.30 g x 1mol/36.5 g = 0.118 moles

Final moles Cl₂ = 0.0130

2HCl_(g) <===>H_2(g) + Cl_2(g) (I'm guessing that this is the actual equation. If not, I have no idea what H₂Cl is)

0.118.............0............0.........Initial

-2x...............+x..........+x.........Change

0.118-2x.......0.0130...0.0130..Equilibrium

From the ICE table x = 0.0130 and moles HCl at equilibrium = 0.118 - 2(0.0130) = 0.118 - 0.026 = 0.092

Equilibrium [HCl] = 0.092 mol/1.5 L = 0.0613 M

a) Kc = [H₂][Cl₂] / [HCl]²

b) [H₂] = [Cl₂] = 0.0130 mol/1.5 L = 0.00867 M

c) Kc = (0.00867)² / 0.0613 = 1.2x10^-3

d) Kp = Kc(RT)^∆n

Kp = 1.2x10^-3 (0.0821*423)⁰

Kp = 4.2x10^-2

e) Use PV = nRT substituting 0.0130 moles of Cl2 for n and 1.5 L for V and solve for P