CH4(g) + H2O(g) // CO(g) + 3H2(g)

Question

Consider the equilibrium system described by the chemical reaction below. At equilibrium, a sample of gas from the system is collected into 5.00 L flask. The flask is found to contain 8.62 g of CO, 2.60 g of H₂, 43.0 g of CH₄, and 48.4 g of H₂O at 320 °C. What are the values of Kc and Kp for this reaction? (33)

Answer 1

Convert all to moles / L

CO = 8.62 g x 1 mol CO/28 g = 0.308 moles CO = 0.0616 M

H2 = 2.60 g H2 x 1 mol H2 / 2 g = 1.30 moles H2 / 5 L = 0.65 M

CH4 = 43.0 g CH4 x 1 mol CH4 / 16 g = 2.69 moles / 5 L = 0.538 M

H2O = 48.4 g H2O x 1 mol H2O/18 g = 2.69 moles / 5 L = 0.538 M

CH4(g) + H2O(g) <==> CO(g) + 3H2(g)

0.538..........0.538............0.0616.......0.65.....Equilibrium concentrations

Kc = [CO][H₂]³ / [CH₄][H₂O] = (0.0616)(0.275) / (0.538)(0.538)

Kc = 0.0585

Kp = Kc(RT)^∆n where ∆n is change in # of moles of gas = 4 moles - 2 moles = 2 moles

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 320 + 273 = 593

Kp = (0.0585)(0.0821* 593)²

Kp = 139