Question

MnO2 (s) + 2 CO (g) → Mn (s) + 2 CO2 (g)

5. Calculate ΔG at 298K if MnO2 = 0.150 mol, Mn = 0.850 mol, CO = 0.250 M and CO2 = 0.500 M in a 1.25 L flask.

Answer 1

MnO₂ (s) + 2 CO (g) → Mn (s) + 2 CO₂ (g)

Concentration of MnO₂ is = [MnO₂] = number of moles / volume in L

                                                    = 0.150 mol / 1.25 L

                                                    = 0.12 M

[Mn ] = 0.850 mol / 1.25 L

       = 0.68 M

Also given that [CO] = 0.250 M

                      [CO₂] = 0.500 M

Equilibrium constant , K= ([Mn][CO₂]²) / ( [MnO₂][CO]²)

                                   = (0.68 x 0.500² ) / ( 0.12 x 0.250²)

                                  = 22.67

We know that ΔG = -RT lnK

Where

R = gas constant = 0.0821 L-atm / (mol-K)

T = temperature = 298 K

Plug the values we get

ΔG = -(0.0821 x 298 ) x ln 22.67

    = -76.35 J