The water-gas reaction is a source of hydrogen... Show work if possible, thanks!

Question

The water-gas reaction is a source of hydrogen...

Show work if possible, thanks!

OAST Tutorial Problem Water Gas The water-gas reaction is a source of hydrogen. Passing steam over hot carbon produces a mixture of carbon monoxide and hydrogen: H2o The value of Ke for the reaction at 1000°C is 3.00 x 10-2 0.442 atm and a. Calculate the equilibrium partial pressures of the products and reactants if PH20 Pco = 5.000 atm at the start of the reaction. Assume that the carbon is in excess. Number Number atm atm Number atm Co b. Determine the equilibrium partial pressures of H20, CO, and H2 after CO and H2 at 0.078 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

0.078 atm of CO and H2 are added

so

pH2 = 0.167 + 0.078 = 0.245

pCO = 5.167 + 0.078 = 5.245

now

using ICE table

at equilirbium

pH2 = 0.245 - x

pCO = 5.245 - x

pH20 = 0.275 + x

now

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

so

3.1354 = [5.245 -x ] [0.245 - x] / [0.275 + x]

0.768173 + 3.1354x = 1.285025 - 5.49x + x2

x2 - 8.6254x + 0.516852 = 0

x = 0.060344

so

pH2 = 0.245 - 0.06 = 0.185 atm

pCO = 5.245 - 0.06 = 5.185 atm

pH20 = 0.275 + 0.06 = 0.335 atm

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