Question

When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion per mole of methane under these conditions?

Short answer

Enthalpy of combustion per mole of methane under these conditions will be -800 kJ.

Step-by-step solution

Number of moles

The amount of heat given out when one mole of a substance is burnt in the presence of excess oxygen, the amount of heat generated is called heat of combustion.

We have to calculate the moles of methane present in 2.50 g of methane.

\(\begin{array}{l}{\rm{No of moles = }}\frac{{{\rm{Weight of substance}}}}{{{\rm{Molecular mass of substance}}}}\\\\ \Rightarrow {\rm{ No of moles of methane = }}\frac{{{\rm{Weight of methane present = }}2.50{\rm{ }}g}}{{{\rm{Molecular mass of methane = }}16{\rm{ }}gmo{l^{ - 1}}}}\\{\rm{ = }}0.15625{\rm{ }}mol\end{array}\)

Enthalpy of combustion

Now, we have to calculate the enthalpy of combustion per mole of methane under standard conditions.

\(\begin{array}{l}{\rm{ Burning 0}}{\rm{.15625 mole of methane, 125kJ of heat is produced}}\\{\rm{On burning 1 mole of methane }}\frac{{{\rm{125kJ \times 1 mole}}}}{{{\rm{0}}{\rm{.15625 mole}}}}{\rm{ of heat is produced}}\\{\rm{ = 800 kJ of heat is produced}}\end{array}\)

Hence, the enthalpy of combustion of methane is =-800kJmol^-1

[Negative sign indicates that energy is released]