Question

Calculate the work (in kJ) when 2.60 moles of methane react with excess oxygen at 363 K:

CH₄(g) + 2O₂(g) ? CO₂(g) + 2H₂O(l)

Answer 1

Work = F*d = P?V

n CH4 = 2.6 moles

n O2 = excess

P1V1 = n1RT1

P2V2 = n2RT2

Problem does not states if P is constant, but this is common for processes involving combustion... Assume constant pressure. P1=P2. Also, the problem states @ T = 363 asume T1=T2

Divide equations 1 and 2

P1V1/ P2V2= n1RT1/ n2RT2

V1/V2 = n1/n2

V1 = n1/n2 * V2

NOTE... the water is condensed so the volume of this water compared to the volume of CO2 gas should be almost 0... then... asume only CO2 is in the vapor phase

NOTE2: No data on excess oxygen is given, suppose all oxygen is consumed and no O2 is left... This is in fact false since we should account all the left over O2 but there is no data to do so... Asume O2 final is = 0.

From the equation

Reactants = 1 mol of CH4:2 mol of O2

2.6 mol : 5.2 mol O2 = 7.8 mol

Products = 1 mol of CH4 :1 mol CO2 : 2 mol of H2O

2.6 mol CH4 -> 2.6 mol of CO2 and 5.2 mol of H2O

Total mol = 7.8 mol

Assume Ideal Gas once again, ignore the H2O in liquid phase

Moles of Reactants = 7.8

Moles of Product = 2.6

Substitute in V1 = n1/n2 * V2

V1 = (7.8/2.6) * V2

NOTE... once again we cant calculate the volume since there is no Pressure given... and no data to calculate either P or V

PV = nRT (we have only n and T, missing P and V)

We can only set an equation for work as follows:

W = P(V2-V1) = P*[V2-(7.8/2.6) * V2]