Using standard thermodynamic data from Appendix L of your text, calculate the equilibrium constant at 298 K for the following chemical reaction:

CO(g) + H2O(l) CO2(g) + H2(g).

Show work please!

Thank you!!!

Follow these directions.

Look up the delta Go f (that's delta Gof of each of the products and each of the reactants. Pure elements like H2 are zero but the others are not.

Then calculate dGo f for the reaction by
dGofrxn = (n*dGf products) - (n*dGf reactants)
Finally, dG = -RTln*K

Then d

To calculate the equilibrium constant (K) for the given chemical reaction, we need to use the standard Gibbs free energy change (ΔG°) for the reaction at 298 K. The relationship between ΔG° and K is given by the equation:

ΔG° = -RT ln(K),

where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (298 K in this case).

To find the value of ΔG°, we will need to use the standard Gibbs free energy of formation (ΔG°f) values for each compound involved in the reaction. Appendix L of your text contains the necessary data.

The chemical equation given is:

CO(g) + H2O(l) → CO2(g) + H2(g).

The standard Gibbs free energy change (ΔG°) for this reaction can be calculated using the formula:

ΔG° = ΣnΔG°f(products) - ΣnΔG°f(reactants),

where Σn represents the sum of the stoichiometric coefficients of each species and ΔG°f is the standard Gibbs free energy of formation.

Let's find the ΔG° values for all the compounds involved:

ΔG°f(CO(g)) = -137.3 kJ/mol
ΔG°f(H2O(l)) = -237.2 kJ/mol
ΔG°f(CO2(g)) = -394.4 kJ/mol
ΔG°f(H2(g)) = 0 kJ/mol

Next, we calculate the ΔG° for the given reaction:

ΔG° = (1 ΔG°f(CO2)) + (1 ΔG°f(H2)) - (1 ΔG°f(CO)) - (1 ΔG°f(H2O))
= (-394.4 kJ/mol) + (0 kJ/mol) - (-137.3 kJ/mol) - (-237.2 kJ/mol)
= -394.4 + 137.3 + 237.2 - (-237.2)
= -173.9 kJ/mol

To convert from kJ to J, multiply by 1000:

ΔG° = -173.9 kJ/mol × 1000 J/kJ
= -173,900 J/mol

Now, we can use the value of ΔG° in the equation to calculate the equilibrium constant (K):

ΔG° = -RT ln(K),
-173,900 J/mol = -(8.314 J/mol·K)(298 K) ln(K)

Now, solve for ln(K):

ln(K) = -173,900 J/mol / (8.314 J/mol·K × 298 K)

ln(K) ≈ -73.65

Finally, calculate K by taking the exponent of both sides:

K = e^(-73.65)

K ≈ 2.49 x 10^(-32)