Question

A mixture of CH4 and H2O is passed over a catalyst at 1000 K. The emerging gas is collected into a 5.00 L flask and 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. The reaction is at equilibrium in the flask. Calculate Kc. The reaction is shown. HINT: calculate the [M] at equilibrium of each compound by converting g to moles and dividing by L. Show your Kc expression. [5pts]

CH4(g) + H2O(g) ⇋ CO(g) + 3 H2(g).

Answer 1

The flask contains:-

8.62 g of CO

2.60 g of H2

43.0 g of CH4

48.4 g of H2O

First, we will convert the given mass of compounds into no. of moles and then using the no. of moles of compound and volume of flask we will find their concentrations:-

Finding the no. of moles of each compound:-

Molecular weight of CO = 28.01 g/mol

i.e., 28.01 g of CO = 1 mole

1 g of CO = (1/28.01) mole

8.62 g of CO = 8.62(1/28.01) moles

= 0.3077 moles

Molecular weight of H2 = 2.01 g/mol

i.e., 2.01 g of H2 = 1 mole

1 g of H2 = (1/2.01) mole

2.60 g of H2 = 2.60(1/2.01) moles

= 1.2935 moles

Molecular weight of CH4 = 16.04 g/mol

i.e., 16.04 g of CH4 = 1 mole

1 g of CH4 = (1/16.04) mole

43.0 g of CH4 = 43(1/16.04) moles

= 2.6808 moles

Molecular weight of H2O = 18.01 g/mol

i.e., 18.01 g of H2O = 1 mole

1 g of H2O = (1/18.01) mole

48.4 g of H2O = 48.4(1/18.01) moles

= 2.6874 moles

Finding the concentration of each compound:-

using the formula,

Concentrationi.e., Molarity (M) = 1 Volumeof flask -----(a)

Volume of flask = 5.0 L -----(Given)

Concentration of CO = (0.3077/5) -----using equation(a)

[CO] = 0.0615 M

Concentration of H2 = (1.2935/5) -----using equation(a)

[H2] = 0.2587 M

Concentration of CH4 = (2.6808/5) -----using equation(a)

[CH4] = 0.5362 M

Concentration of H2O = (2.6874/5) -----using equation(a)

[H2O] = 0.5375 M

Balanced chemical equation at equilibrium:-

CH4(g) + H2O(g) ⇋ CO(g) + 3 H2(g)

The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.

Therefore, the expression for Kc can be written as,

Kc = CH., 4.0 -----(6)

putting the concentrations of each compound in equation(b), we get,

_ 0.0615 0.2587 Kc = 10.5362 0.5375

Kc = 10.5362 0.5375) K _ 0.0615 0.0173

Kc = 0.0011 (0.2882

Kc = 0.0038
The value of Kc for the given reaction at equilibrium is 0.0038