Question

1. Enthalpies and energies of reactions. Consider the combustion of methane, which involves the reaction of 1 mol CH4(g) with 2 mol of O2(g) to form 1 mol CO2 (g) and 2 mol H2O (l): CH4(g) + 2 O2(g) → CO2(g)+ 2 H2O(l) (a) Using heats of formation from Table 2.6 in the text (or from the tables in the back of the book, or from webbook.nist.gov) calculate ∆Hr◦ for the reaction. (b) Calculate ∆Ur for the reaction by only taking into account the gaseous products and reactants. (c) Now, calculate ∆Ur for the reaction taking into account the volume of water as well. Is it a good approximation to ignore the volume of the water?

Answer 1

Reaction is

CH4 + 2O2 --> CO2 + 2H2O

ΔHrxn = [ΔH_CO2 + 2 X ΔH_H2O] - [Δ_CH4 + ΔH_O2]

ΔHrxn = [ -393.5+ 2 X -285.8] - [-74.8 + 0]   KJ/ mole

ΔHrxn =-965.1 + 74.8 = -890.3 KJ / mole

We know that

ΔHrxn = ΔU + ΔnRT

where

Δn = 1- 3 = -2

R = 8.314 J / mol K = 8.314 X 10^-3 KJ / mol K

T = 273.15 K [as we study combustion at STP]

ΔH - ΔnRT = ΔU

ΔU = -890.3 - (-2) (8.314 X 10^-3 X 273.15] = -885.75 KJ / mol

c) if we consider the volume of water also then

Δn = 0

Hence ΔH = ΔU = -890.3 KJ / mol

So it is good approximation to ignore the volume of water