Question

Calculate the work (in kJ) when 2.70 moles of methane react with excess oxygen at 364 K: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

Answer 1

Ans. Number of moles of gases before combustion-

            CH₄ = 2.7 moles

            O₂ = X moles

Total initial moles = 2.70 mol + X mol

# Change in number of moles of gases during combustion-

            Moles of CH₄ consumed = 2.7 moles

            Moles of O₂ consumed = 2 x moles of CH₄ = 2 x 2.70 mol = 5.40 mol

                        Moles of O₂ remaining = Initial moles – Moles of O₂ consumed

                                                = (X moles – 5.40 mol)

            Moles of CO₂ formed = moles of CH₄ = 2.70 mol

Since H₂O formed is in liquid form, its assumed that H₂O(l) contributes negligible volume when compared to the volume of gases.

# Moles of gases remaining after combustion-  

Total moles remaining = Moles of CO₂ formed + Moles of O₂ remaining

                                    = 2.70 mol + (X moles – 5.40 mol)

                                    = X mol – 2.70 mol

# Change in number of moles of gases during combustion-

                        Total moles of after combustion – Total moles before combustion

                        = (X mol – 2.70 mol) – (X mol – 2.70 mol)

                        = 0

# Following Avogadro’s Law equal volume of all gases have equal number of moles, at constant pressure and temperature.

                        n1V1 = n2V2                         -P, T constant

Since change in number of moles of gaseous chemical species is zero (0), the change in volume due to combustion would also be zero because volume is proportional to number of moles.

            That is, change in volume, dV = 0

# Now,

Work done, W = p (dV)                  - at constant pressure

                        Or, W = p x 0.0

                        Hence, W = 0.0

Hence, work done = 0.0 kJ