Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.68 moles of CO(g) react at standard conditions.

Question

Consider the reaction

CO(g) + H₂O(l)CO₂(g) + H₂(g)

Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.68 moles of CO(g) react at standard conditions.

    S°_system =   J/K

Answer 1

Here are the standard absolute entropies I found online. You may want to double check with your textbook to make sure these values are consistent:

CO(g) = 197.9 J/molK

H₂O(g) = 188.7 J/molK

CO₂(g) = 213.8 J/molK

H₂(g) = 130.5 J/molK

dS_rxn = S_products - S_reactants

CO(g) + H₂O(l)CO₂(g) + H₂(g)

1.) Multiply each S⁰ by the stoichiometric coefficient and solve for dS_rxn

dS_rxn = [(1 mol CO₂)(213.8 J/molK) + (1 mol H₂)(130.5 J/molK)] - [(1 mol CO)(197.9 J/molK) + (1mol H₂O)(188.7 J/molK)] = (213.8 J/K + 130.5 J/K) - (197.9 J/K + 188.7 J/K) = -42.3 J/molK rxn

2.) Given 1.68 mol CO(g) reacted, find dS_sys using the stoichiometric ratio

dS_sys = -42.3 J/molK --> 1 mol CO(g) (from the balanced equation)

dS_sys = -42.3 J/molCO_(g)K x 1.68 mol CO(g) ...moles should cancel leaving you with J/K

dS_sys = -71.1 J/K

Hope this helps!