Calculate the equilibrium constant at 300 K

for the reaction
HgO(s) ↽⇀ Hg(ℓ) +
1/2O2(g),
given the data

delta S delta h
� kJ
J/K mol mol

HgO(s) 70.29 −90.83
Hg(ℓ) 76.02 0
O2(g) 205.14 0

I can't tell what your columns/numbers mean.

dH rxn = (n*dH products) - (n*dH reactants). Solve for dH rxn.
dS rxn = (n*dS products) - (n*dS reactants). Solve for dS rxn.
dG = dH - TdS. Solve for dG.
Then dG = -RT*lnK. Solve for K. T must be in kelvin, use 8.314 for R.

To calculate the equilibrium constant (K) at 300 K for the given reaction, we need to use the equation:

ΔG° = -RT ln(K)

where:
- ΔG° is the standard Gibbs free energy change,
- R is the gas constant (8.314 J/(K mol)),
- T is the temperature in Kelvin, and
- ln denotes the natural logarithm.

First, we need to calculate the standard Gibbs free energy change for the reaction (ΔG°). ΔG° can be calculated using the equation:

ΔG° = ΔH° - TΔS°

where:
- ΔH° is the standard enthalpy change,
- ΔS° is the standard entropy change, and
- T is the temperature in Kelvin.

Given data:
HgO(s): ΔS° = 70.29 J/K mol, ΔH° = -90.83 kJ
Hg(ℓ): ΔS° = 76.02 J/K mol, ΔH° = 0 kJ
O2(g): ΔS° = 205.14 J/K mol, ΔH° = 0 kJ

Now, convert the units of ΔH° from kJ to J by multiplying by 1000. ΔH° for HgO(s) becomes -90.83 kJ * 1000 = -90830 J.

Calculate ΔG° as follows:

ΔG° = ΔH° - TΔS°
= -90830 J - T * (70.29 J/K mol - 76.02 J/K mol - 0.5 * 205.14 J/K mol)

Substitute T = 300 K and calculate ΔG°.

ΔG° = -90830 J - 300 K * (70.29 J/K mol - 76.02 J/K mol - 0.5 * 205.14 J/K mol)

Now, substitute ΔG° into the equation for ΔG° = -RT ln(K), and solve for K:

-RT ln(K) = ΔG°
ln(K) = -ΔG° / (RT)
K = e^(-ΔG° / (RT))

Substitute the values for R, T, and ΔG°, and calculate K:

R = 8.314 J/(K mol)
T = 300 K
ΔG° = calculated ΔG° value

K = e^(-ΔG° / (RT))

After solving this equation, you will obtain the equilibrium constant (K) at 300 K for the given reaction.