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Carbon dioxide and water react to form methanol and oxygen, like this:

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

At a certain temperature, a chemist finds that a 7.5L reaction vessel containing a mixture of carbon dioxide, water, methanol, and oxygen at equilibrium has the following composition:

Compound Amount
CO2 3.28g
H2O 3.86g
CH3OH 1.51g
O2 2.80g

Required:
Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

Sagot :

Maacastrobridn

Answer:

24x10³

Explanation:

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

The equilibrium constant for this reaction is:

Kc = [tex]\frac{[O_2]^3}{[CO_2]^2[H_2O]^4}[/tex]

The expression of [CH₃OH] is left out as it is a pure liquid.

Now we convert the given masses of the relevant species into moles, using their respective molar masses:

  • CO₂ ⇒ 3.28 g ÷ 44 g/mol = 0.0745 mol CO₂
  • H₂O ⇒ 3.86 g ÷ 18 g/mol = 0.214 mol H₂O
  • O₂ ⇒ 2.80 g ÷ 32 g/mol = 0.0875 mol O₂

Then we calculate the concentrations:

  • [CO₂] = 0.0745 mol / 7.5 L = 0.0099 M
  • [H₂O] = 0.214 mol / 7.5 L = 0.0285 M
  • [O₂] = 0.0875 mol / 7.5 L = 0.0117 M

Finally we calculate Kc:

  • Kc = [tex]\frac{0.0117^3}{0.0099^2*0.0285^4}[/tex] = 24x10³

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