Question

Methanol and oxygen react to form carbon dioxide and water, like this: 2CH3OH (l) + 3O2 (g) → 2CO2 (g) + 4H2O (g) At a certain temperature, a chemist finds that a 9.3L reaction vessel containing a mixture of methanol, oxygen, carbon dioxide, and water at equilibrium has the following composition:compound CH3OH O2 CO2 H2Oamount 1.47g 1.56g 2.28g 3.33gCalculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

Answer 1

Answer:

The value of the equilibrium constant Kc for this reaction is 3.34*10^-5

Explanation:

Step 1: Data given

Volume of the reaction vessel = 9.3 L

Mass of CH3OH at equilibrium = 1.47 grams

Mass of O2 at equilibrium = 1.56 grams

Mass of CO2 at equilibrium = 2.28 grams

Mass of H2O at equilibrium = 3.33 grams

Molar mass CH3OH = 32.04 g/mol

Molar mass of O2 = 32.0 g/mol

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

2CH3OH (l) + 3O2 (g) → 2CO2 (g) + 4H2O (g)

Step 3: Calulate moles

Moles = mass / molar mass

Moles CH3OH = 1.47 grams / 32.04 g/mol

Moles CH3OH = 0.0459 moles

Moles O2 = 1.56 grams / 32.0 g/mol

Moles O2 = 0.0489 moles

Moles CO2 = 2.28 grams / 44.01 g/mol

Moles CO2 = 0.0518 moles

Moles H2O = 3.33 grams / 18.02 g/mol

Moles H2O = 0.185 moles

Step 4: Calculate molarity

Molarity = moles / volume

[CH3OH] = 0.0459 moles / 9.3 L

[CH3OH] = 0.00494 M

[O2] = 0.0489 moles / 9.3 L

[O2] = 0.00526 M

[CO2] = 0.0518 moles / 9.3 L

[CO2] = 0.00557 M

[H2O] = 0.185 moles / 9.3 L

[H2O] = 0.0199 M

Step 5: Calculate Kc

Since liquids and solids do not affect the equilibrium, CH3OH will not take part

Kc = [CO2]²[H2O]^4 / [O2]³

Kc = (0.00557² * 0.0199^4) / 0.00526³

Kc = 3.34*10^-5

The value of the equilibrium constant Kc for this reaction is 3.34*10^-5