Question

A mixture of 0.008603 mol of Cl₂, 0.05744 mol of H₂O, 0.05353 mol of HCl, and 0.06453 mol of O₂ is placed in a 1.0-L steel pressure vessel at 1529 K. The following equilibrium is established:

2 Cl₂(g) + 2 H₂O(g) 4 HCl(g) + 1 O₂(g)

At equilibrium 0.006795 mol of Cl₂ is found in the reaction mixture.





(a) Calculate the equilibrium partial pressures of Cl₂, H₂O, HCl, and O₂.


P_eq(Cl₂) =  .

P_eq(H₂O) =  .

P_eq(HCl) =  .

P_eq(O₂) =  .



(b) Calculate K_P for this reaction.


K_P =  .

Answer 1

(a)

The given equilibrium is

2 C12(g) + 2 H2O(g) = 4 HCl(g) +1 O2(g)

Given the initial number of moles of each gas, we can create the following ICE table

	2 Cl2(g)	+2 H2O(g)	4HCG	+1 029)
Initial, mol	0.008603	0.05744	0.05353	0.06453
Change, mol	-2x	-2x	+4x	+x
Equilibrium, mol	0.008603-2x	0.05744 - 2x	0.05353+4x	0.0643 + x

Now, it is given that at equilibrium, the number of moles of Cl2 gas is 0.006795 mol.

Hence, from the ICE table,

ni = 0.008603 mol - 2.x = 0.006795 mol 2.r = 0.008603 mol – 0.006795 mol = 0.001808 mol I=- 0.001808 mol _ 0.000904 mol

It is given that the volume of the container =V= 1.0 L

Temperature , T = 1529 K

Assuming that the gases obey ideal gas relations, their partial pressures can be calculated by using ideal gas equation

nRT P=

Hence, we can find the equilibrium moles and equilibrium partial pressures of each gas as follows:

Equilibrium moles of Cl2 = 0.006795 mol

Hence,

nai, RT K-1 mol-1 x 1529 K Pleq),Cl2 = 0.006795 mol x 0.082057 L atm 1L > Pleq),Cl2 = 0.8525 atm

Equilibrium moles of H2O can be calculated as

nh,0 = 0.05744 mol-2.1 = 0.05744 mol-2x0.000904 mol = 0.055632 mol

Hence,

0.055632 mol x 0.082057 L atm K-1 mol-1 x 1529 K nh,0RT Pleg),420 = → Pleg), H20 = 6.680 atm 1 L

Equilibrium moles of HCl can be calculated as

nuci = 0.05353 mol+4x = 0.05353 mol+4x0.000904 mol = 0.057146 mol

Hence,

0.057146 mol x 0.082057 L atm K-1 mol-1 x 1529 K nhCRT Pleg),HCI = Pleg), HCI = 7.170 atm 11

Equilibrium moles of O2 can be calculated as

no, = 0.06453 mol +1 = 0.06453 mol + 0.000904 mol = 0.065434 mol

Hence,

no, RT Pleg,02 =- Pleg,02 = 8.210 atm 0.065434 mol x 0.082057 L atm K-1 mol-1 x 1529 K 1 L

(b)

Hence, Kp of the reaction can be calculated as follows:

Ple),Hci * Pleg),O2 Pleg).clX P .H (7.170)" x 8.210 0.85252 x 6.6802 669

Hence, the Kp of the reaction is about 669.