Chem HW Problem

Question

A 2.50 L flask initially contains 5.0 g of methane and 10.0 g of oxygen. How many grams of each species (including water) will be present when the reaction reaches equilibrium at 25°C.

CH4(g) + 2O2(g) ⇌ CO2(g) + 2H2O(l)

At equilibrium:

Mass of CH4 =

Mass of O2 =

Mass of CO2 =

Mass of H2O =

Potentially Useful Information:

2H2(g) + O2(g) → 2H2O(l) ΔG° = −47.4 kJ

C(s) + O2(g) → CO2(g) ΔG° = −39.4 kJ

2H2(g) + C(s) → CH4(g) ΔG° = −5.1 kJ

E = mc2

Answer 1

Use Hess' Law to find the ∆G for the reaction:

CH4(g) ==> 2H2(g) + C(s) ... ∆G = +5.1 kJ

2H2(g) + O2(g) => 2H2O(l) ... ∆G = -47.4 kJ

C(s) + O2(g) => CO2(g) ... ∆G = -39.4 kJ

CH4(g) + 2H2 + O2 + C + O2 ==> 2H2 + C + 2H2O + CO2

CH4(g) + 2O2(g) ==> CO2(g) + 2H2O ... ∆G = 5.1 kJ - 47.4 kJ - 39.4 kJ = -81.7 kJ

Now we can use ∆G = -RT ln K to find the equilibrium constant (remember to use units of R and G that agree)

-81700 J = -8.314 J/kmol x 298K ln K

ln K = 32.98

K = 2.1x10¹⁴

CH4(g) + 2O2(g) ⇌ CO2(g) + 2H2O(l)

0.125..... . 0.125.......... .. ..0........... ...0........Initial

-x..............-2x...................+x............+2x....................Change

0.125-x..0.125-2x............x................2x..............Equilibrium

K = 2.1x10¹⁴ = (x)(2x)/(0.125-x)(0.125-2x)

Solve for x and that will be [CO2]. [H2O] will be twice that and [CH4] = 0.125-x and [O2] = 0.125-2x.

Then convert concentrations to mass using the molar mass of each species.