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tiny-mole [99]
3 years ago
13

In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.778 L

flask at 1,072 K. At equilibrium, the flask contains 0.206 mol of CO gas, 0.187 mol of H2 gas, and 0.187 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1,072 K)? Enter to 4 decimal places. HINT: Look at sample problem 17.7 in the 8th ed Silberberg book. Write a balanced chemical equation. Write the Kc expression. Calculate the equilibrium concentrations of all the species given (moles/liter). Put values into Kc expression, solve for the unknown.
Chemistry
1 answer:
sergejj [24]3 years ago
7 0

Answer:

the water concentration at equilibrium is

⇒ [ H2O(g) ] = 0.0510 mol/L

Explanation:

  • CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)

∴ Kc = ( [ CO(g) ] * [ H2 ]³ ) / ( [ CH4(g) ] * [ H2O(g) ] ) = 0,30

  • equilibrium:

⇒ [ CO(g) ] = 0.206 mol / 0.778 L = 0.2648 mol/L

⇒ [ H2(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L

⇒ [ CH4(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L

replacing in Kc:

⇒ ((0.2648) * (0.2404)³) / ([ H2O(g) ] * 0.2404 ) = 0.30

⇒ 0.0721 [ H2O(g) ] = 3.679 E-3

⇒ [ H2O(g) ] = 0.0510 mol/L

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Answer: The enthalpy change is 34.3 kJ

Explanation:

The conversions involved in this process are :

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Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})]

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c_{l} = specific heat of liquid water = 4.184J/g^0C

n = number of moles of water = \frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{72.0g}{18g/mole}=4.00moles

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Now put all the given values in the above expression, we get

\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]\Delta H=34279.8J=34.3kJ        (1 KJ = 1000 J)

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