Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15 K.

Question

Consider the reaction

S(s,rhombic) + 2CO(g)SO₂(g) + 2C(s,graphite)

Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15 K.

ANSWER:  

Answer 1

You are asked to find the equilibrium constant given the reaction:

S(s,rhombic) + 2CO(g)SO₂(g) + 2C(s,graphite)

The equilibrium constant (K) and dG⁰_rxn can be related with the following equation:

dG⁰_rxn = -RT lnK

So first we must solve for dG⁰_rxn:

dG⁰_rxn = dG_products - dG_reactants

I found the following values for dG⁰ but you may want to check with your textbook to make sure these are consistent

S(s,rhombic) = 0 kJ/mol

CO(g) = -137.17 kJ/mol

SO₂(g) = -300.19 kJ/mol

C(graphite) = 0 kJ/mol

In order to calculate dG⁰_rxn we must multiply these values by their stoichiometric coefficients

dG⁰_rxn = dG_products - dG_reactants

dG⁰_rxn = [(2 mol C_(s))(0 kJ/mol) + (1 mol SO_2(g))(-300.19 kJ/mol)] - [(2 mol CO_(g))(-137.17 kJ/mol) + (1 mol S_(rhombic))(0kJ/mol)] = (0 kJ - 300.19 kJ) - (-274.34 kJ + 0 kJ) = -300.19 kJ + 274.34 kJ = -25.85 kJ/mol rxn

Now find the equilibrium constant (K) using the equation below, where R is the gas constant (8.314 J/molK) and T is temperature (given as 298.15 K):

dG⁰_rxn = -RT lnK

We can rearrange this formula to find K ...(don't forget to convert J to kJ in order to make all units cancel!):

lnK = dG⁰_rxn / -RT

lnK = (-25.85 kJ/mol) / -(8.314 J/molK)(298.15 K)(1 kJ/10³ J) ...J, mol and K should cancel

lnK = 10.4

K = e^10.4 = 3.38 x 10⁴