Making Hydrogen Gas Passing steam over hot carbon produces a mixture of carbon m

ID: 947342 • Letter: M

Question

Making Hydrogen GasPassing steam over hot carbon produces a mixture of carbon monoxide

and hydrogen:

H2O(g) + C(s) <=> CO(g) + H2(g)

The value of Kc for the reaction at 1000°C is 3.0 × 10–2.

a. Calculate the equilibrium partial pressures of the products and reactants if PH2O = 0.442 atm

and PCO = 5.0 atm at the start of the reaction. Assume that the carbon is in excess.

b. Determine the equilibrium partial pressures of the reactants and products after sufficient CO

and H2 are added to the equilibrium mixture in part a to initially increase the partial pressures

of both gases by 0.075 atm.

a. Kc = [CO][H2]/[H2O]

let x be the change at equilibrium

3.0 x 10^-2 = (5 + x)(x)/(0.442 - x)

0.01326 - 3 x 10^-2x = 5x + x^2

x^2 + 5.03x - 0.01326 = 0

x = 2.63 x 10^-3

Thus equilibrium partial pressure for,

[H2O] = 0.442 - 2.63 x 10^-3 = 0.4394 atm

[CO] = 5 + 2.63 x 10^-2 = 5.00263 atm

[H2] = 2.63 x 10^-2 atm

b. with,

[CO] = 5.00263 + 0.075 = 5.07763 atm

[H2] = 2.63 x 10^-3 + 0.075 = 0.007763 atm

Kc = 3 x 10^-2 = (5.00763 - x)(0.007763 - x)/(0.4394 + x)

0.0132 + 3 x 10^-2x = 0.039 - 5.0154x + x^2

x^2 - 5.0454x + 0.0258 = 0

x = 5.12 x 10^-3 atm

Thus equilibrium partial pressure for,

[H2O] = 0.4394 + 5.12 x 10^-3 = 0.4445 atm

[CO] = 5.07763 - 5.12 x 10^-3 = 5.0725 atm

[H2] = 0.007763 - 5.12 x 10^-3 = 2.643 x 10^-3 atm

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