Question

Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The equilibrium constant for the reaction at 700.0 K is Kp=1.60×10−3. If a 1.55-L reaction vessel initially contains 247 torr of water at 700.0 K in contact with excess solid carbon, find the percent by mass of hydrogen gas of the gaseous reaction mixture at equilibrium.

Answer 1

The reaction proceed according to the reaction equation:
C(s) + H₂O(g) ⇄ CO(g) + H₂

In equilibrium state the reaction quotient in terms of activities equals equilibrium constant, i.e.
K = (a_CO ∙ a_H₂) / (a_C ∙ a_H₂O)

Assuming ideal gas mixture the activity of the gaseous components equals their mole fraction x_i in the gas phase. Carbon is present in a separate phase as pure solid. So it has activity
a_C = 1
Hence equilibrium equation can be rewritten as:
K = (x_CO ∙ x_H₂) / x_H₂O

Next consider the number of moles of the gaseous components to determine their mole fractions Since carbon has a much higher density than the gases, we can neglect is small volume and its change in volume due to reaction and approximate the whole reactor filled with gas. Assuming ideal gas behavior you can compute initial amount of water vapor in the vessel from ideal gas law:
n₀ = P∙V / (R∙T)
     = 249 ∙ (101325/760) Pa∙ 1.55×10⁻³m³ / (8.3145 Pa∙m³∙K⁻¹∙mol⁻¹ ∙ 700K)
     = 8.841×10⁻³ mol

Let y be the amount of water, which has reacted away as equilibrium has established. Then amount of water in equilibrium mixture is:
n_H₂O = n₀ - y
According to reaction one mole of each hydrogen and carbon dioxide is formed per mole of water consumed, that means theirs amounts in the equilibrium mixture are:
n_CO = n_H₂ = y
The total number of moles in equilibrium is:
n_t = n_H₂O + n_CO + n_H₂ =
= n₀ - y + y + y
= n₀ + y

So the mole fraction in the equilibrium mixture are:
x_H₂O = n_H₂O / n_t = (n₀ - y)/(n₀ + y)
x_CO = n_CO / n_t = y/(n₀ + y)
x_H₂ = n_H₂ / n_t = y/(n₀ + y)

By substituting these mole fractions to the equilibrium relation you get:
K = [y/(n₀ + y)]² / [(n₀ - y)/(n₀ + y)]
<=>
K = y² / [(n₀ - y)∙(n₀ + y)]
<=>
K = y² / (n₀² - y²)
<=>
K∙(n₀² - y²)∙ = y²
<=>
y²∙(K + 1) = n₀²∙K
=>
y = n₀∙√( K/(K + 1) )
= 8.841×10^-3 mol ∙ √( 1.60×10^-3 / (1.60×10^-3 + 1) )
= 3.534×10^-4 mol

So the composition of the equilibrium mixture is:
x_CO = x_H₂ = y/(n₀ + y)
= 3.534×10^-4 mol / (8.841×10^-3 mol + 8.841×10^-3 mol)
= 0.01999
= 1.999 %
x_H₂O = (n₀ - y)/(n₀ + y)
= (8.841×10^-3 mol + 3.534×10^-4 mol)/(8.841×10^-3 mol + 8.841×10^-3mol)
= 0.52
= 52 %