Question

An industrial chemist introduces 1.6 atm H₂ and 1.6 atm CO₂ into a 1.00-L container at 25.0°C and then raises the temperature to 700.0°C, at which K_eq = 0.534:
H₂(g) + CO₂(g) ⇔ H₂O(g) + CO(g)
How many grams of H₂ are present after equilibrium is established?

Answer 1

Ans. Part A: Given,

            Volume of the reaction vessel = 1.0 L

            Pressure of each gas (individually) = 1.6 atm.

            Temperature, T = 25⁰C = 298 K

Using ideal gas equation Ideal gas Law: pV = nRT           - equation 1

            Where, p = pressure in atm

            V = volume in L

            n = number of moles =

            R = universal gas constant= 0.082057338 atm L mol^-1K^-1

            T = absolute temperature =

Putting the values in equation 1-

            n = pV / RT = (1.6 atm. X 1.0 L) x (0.082057338 atm L mol^-1K^-1 x 298 K)

            Or, n = 0.065431

Thus, number of moles of H₂ = 0.065431 moles

                Also, number of moles of CO₂ = 0.065431 moles

Part B:

Stoichiometry : 1 mol H2 reacts with 1 mol CO2 to produce 1 mol H2O and 1 mol CO.

                                H2(g)                     + CO2 <----------> H2O(g)               +     CO2(g)

Initial >                 0.065431           0.065431                       0                                  0

Change >               -x                     -x                              +x                                    +x          

Equilibrium>     (0.065431 -x)      (0.065431 -x)              +x                                    +x          

Because the volume of vessel is 1.0 L, the concentration of the reactions is equal to number of moles per Litre. So, the concentration of any species is equal to their respective moles.

Now,

Equilibrium constant, Keq = ([CO2] [H2O]) / ([H2] [CO2])

Or, 0.534 = x² / (0.065431 -x)² = x / (0.065431 -x)

Or, x = 0.534 x (0.065431 -x) = 0.034940154 – 0.534 x

Or, x+ 0.534 x = 0.034940154

Or, 1.534 x = 0.034940154

Hence, x = 0.02277

So, [H2] at equilibrium = (0.065431 -x) = 0.065431 – 0.02277 = 0.042661

Moles of H₂ = 0.042661 moles                ; [note: volume = 1 .0 L]

Mass of H2 = moles x molar mass

                        = 0.042661 moles x 2.0 g mol^-1

                        = 0.085322 g

Thus, mass of H₂ remaining at equilibrium = 0.085322