Question

When a mole of methane is burnt, 890KJ of heat is given off at STP .If the standard enthalpies of formation of CO2 and liquid water are -393KJ/mol and -286KJ/mol respectively, what is the standard enthalpy of formation of methane?

Answer 1

To determine the standard enthalpy of formation of methane (CH4), we can use Hess's law, which states that the change in enthalpy for a reaction is independent of the pathway taken. We'll use the following equation to calculate the enthalpy of formation:

CH4 + 2O2 → CO2 + 2H2O

According to the equation, the combustion of methane produces one mole of carbon dioxide (CO2) and two moles of water (H2O). The given values are:

ΔHf°(CO2) = -393 kJ/mol (standard enthalpy of formation of CO2) ΔHf°(H2O) = -286 kJ/mol (standard enthalpy of formation of liquid water) ΔH° = -890 kJ (heat given off during combustion of 1 mole of CH4 at STP)

Now, let's calculate the standard enthalpy of formation of methane (ΔHf°(CH4)):

ΔH° = Σ(ΔHf°(products)) - Σ(ΔHf°(reactants))

-890 kJ = [1 × ΔHf°(CO2) + 2 × ΔHf°(H2O)] - ΔHf°(CH4)

Substituting the given values:

-890 kJ = [1 × (-393 kJ/mol) + 2 × (-286 kJ/mol)] - ΔHf°(CH4)

Simplifying the equation:

-890 kJ = -393 kJ/mol - 572 kJ/mol - ΔHf°(CH4)

-890 kJ = -965 kJ/mol - ΔHf°(CH4)

Rearranging the equation to solve for ΔHf°(CH4):

ΔHf°(CH4) = -965 kJ/mol + 890 kJ

ΔHf°(CH4) = -75 kJ/mol

Therefore, the standard enthalpy of formation of methane (CH4) is -75 kJ/mol.