Question

In a laboratory experiment 24.239 g of methane is burned in air containing O2 to form gaseous CO2 and H2O. Calculate the final temperature in °C of the product mixture if the methane and air are both at an initial temperature of 1.3°C. Assume a stoichiometric ratio of methane to oxygen from the air, with air being 21% O2 by volume and the rest of the volume being N2 (c of CO2 = 57.2 J/molK; c of H2O(g) = 36.0 J/molK; c of N2 = 30.5 J/molK). Use data of formation enthalpies from the appendix if needed.

Enthalpies: CO2(g): -393.5 kJ/ mol, H2O(g): -241.826 kJ/ mol, O2(g): 0 kJ/ mol, N2(g): 0 kJ/ mol, CH4(g): -74.87

Answer 1

The balanced equation for the combustion of methane can be written as follows.

CH₄(g) + 2O₂(g) CO₂(g) + 2H₂O(g)

The no. of moles of CH₄ = 24.239 g/16 g.mol^-1 = 1.51494 mol

The enthalpy of the reaction can be calculated as follows.

H_rxn = {(H)_CO2(g) + 2*(H)_H2O(g)} - {(H)_CH4(g) + 2*(H)_O2(g)}

= (-393.5 + 2*-241.826) - (-74.87 + 2*0)

= -802.282 kJ/mol

Therefore, the heat released in the reaction (q) = 802.282 kJ/mol * 1.51494 mol = 1215.4 kJ

Now, q = m * c * T

Where, 'm' is the mass of methane, 'c' is the specific heat capapcity of methane = 2.191 kJ/kg.K, T = final temperature = initial temperature

Therefore, 1215.4*1000 J = 24.239 g * 2.2 J/g.^oC * (final temperature - 1.3) ^oC

i.e. Final temperature = 22793.3 ^oC