Question

When the oxide of generic metal M is heated at 25.0 °C, only a negligible amount of M is produced.When this reaction is coupled to the conversion of graphite to carbon dioxide, it becomes spontaneous. What is the chemical equation of this coupled process? Show that the reaction is in equilibrium, include physical states, and represent graphite as C(s).What is the thermodynamic equilbrium constant for the coupled reaction? MO(s)<--->M(s)+O2(g)

delta G=288 kJ/mol

Answer 1

Write the two equations.

MO₂ (s) -----> M (s) + O₂ (g)

C (s) + O₂ (g) ---> CO₂ (g)

Add the two equations

MO₂ (s) + C (s) + O₂ ------> M + O₂ + CO₂

O₂ is on each side of the equation so cancel it out.

Then balanced equation is

C(s) + MO₂ (s) ----> M(s) + CO₂ (g)

MO(s)<--->M(s)+O₂(g) ΔG =288 KJ/mol

Now calculate ΔG for reaction

C(s) + MO₂ (s) ----> M(s) + CO₂ (g)

ΔG = ΔG [CO₂]+ ΔG[M(s)]- ΔG [MO_{2 (s)}]- ΔG[C (s)]

ΔG for CO₂=-394.4 KJ/mol

ΔG for a solid is zero

So, ΔG for system = -394.4+288.9

                                 = -105.5

The thermodynamic equilibrium constant for the coupled reaction is calculated by equation

K=e^-((ΔG )/(RT))

Where

T = 25.0 °C = 298 K

R = 8.314 J/mol*K = 8.314/1000 K J/mol*K

K=e^ - ((ΔG)/((8.314/1000)*298)))

K =e^ - ((-105.5)/((8.314/1000)*298)))

K = e^ (42.58)

K = 3.11 x 10^18

The large value for K indicates that reaction is spontaneous.