ΔG° = ΔH° - TΔS°

Question

For the reaction

C₂H₄(g) + H₂O(g)  CH₃CH₂OH(g)

G° = -7.9 kJ and S° = -125.7 J/K at 300 K and 1 atm.

This reaction is (reactant, product) ______favored under standard conditions at 300 K.

The standard enthalpy change for the reaction of 2.36 moles of C₂H₄(g) at this temperature would be____  kJ.

Answer 1

If G is negative, the reaction is thermodynamically favorable in the forward direction (product favored).

If G is positive, the reaction is thermodynamically unfavorable in the forward direction (reactant favored).

To solve for enthalpy, solve for H in the equation:

G=H-TS

Convert S to kJ since G is in kJ

-7.9=H-300(-0.1257)

Solve for H, unit will be in kJ.

Then, since H is in kJ/mol, just multiply by 2.36 moles, like this:

(2.36 mol ) * (x kJ / 1 mol)= kJ

Make sure the kJ answer has the correct sign, positive or negative.