Water Gas The water-gas reaction is a source of hydrogen. Passing steam over hot

Question

Water Gas The water-gas reaction is a source of hydrogen. Passing steam over hot carbon produces a mixture of carbon monoxide and hydrogen: The value of K_c for the reaction at 1000degreeC is 3.00 times 10^-2. Calculate the equilibrium partial pressures of the products and reactants if P_h2o = 0.442 atm and P_co = 5.000 atm at the start of the reaction. Assume that the carbon is in excess. ibrium partial pressures of H_2O, CO, and H_2 after CO and H_2 at 0.072 atm are added ure in part a. Determine the equilibrium partial pressures of H_2O, CO, and H_2 after CO and H_2 at 0.072 atm are added to the equilibrium mixture in part a.

Explanation / Answer

we know that

Kp = Kc (RT)^dn

dn = moles of gases in products - moles of gases in reactants

so

in this case

dn = 1 + 1 - 1 = 1

so

Kp = Kc (RT)

Kp = 3 x 10-2 x 0.0821 x 1273

Kp = 3.1354

now

initially

pH20 = 0.442

pCO = 5

now

H20 + C (s) ---> CO + H2

using ICE table

at equilibrium

pH20 = 0.442 - x

pC0 = 5 + x

pH2 = x

now

Kp = [pCO] [pH2] / [pH20]

so

3.1354 = [5 + x] [x ] / [0.442 - x]

1.3858468 - 3.1354x = 5x + x2

x2 + 8.1354x -1.3858468 = 0

x= 0.167

so

pH20 = 0.442 - 0.167 = 0.275 atm

pCO = 5 + 0.167 = 5.167 atm

pH2 = 0.167 atm

2)

now

0.072 atm of CO and H2 are added

so

pH2 = 0.167 + 0.072 = 0.239

pCO = 5.167 + 0.078 = 5.239

now

using ICE table

at equilirbium

pH2 = 0.239 - x

pCO = 5.239 - x

pH20 = 0.275 + x

now

Kp = [pCO] [pH2] / [pH20]

so

3.1354 = [5.239 -x ] [0.239 - x] / [0.275 + x]

0.768173 + 3.1354x = 1.252121 - 5.478x + x2

x2 - 8.6134x + 0.483948 = 0

x = 0.05655

so

pH2 = 0.245 - 0.05655 = 0.18845 atm

pCO = 5.245 - 0.05655 = 5.18845 atm

pH20 = 0.275 + 0.05655 = 0.33155 atm

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