# Chem HW 9 - Rearrange the equation above to solve for ln...

This preview shows pages 1–3. Sign up to view the full content.

Rearrange the equation above to solve for ln K eq : lnK = - ( G o ) = - ( 83600 )J mol K 1 = -33.726 RT mol 8.314 J 298.15 K Note that all of the units cancel as they should to take the logarithm. Raise to the base "e" to clear the logarithm: K = e -33.726 = 2.25E-15 The change in the Gibbs Free Energy is related to the change in the enthalpy and the entropy. This is summarized by the Gibbs-Helmholtz equation: G o rxn = H o rxn - T S o rxn For the reaction as written at 336 K: G o = H o - T S o = -851.5 kJ - 336 K -37.0 J 1 kJ = -839.1 kJ K 1000 J For the reaction of 1.87 moles of Fe 2 O 3 (s): G o = 1.87 mol Fe 2 O 3 (s) -839.1 kJ = -1.57E+3 kJ 1 mol Fe 2 O 3 (s) Since G o is less than zero, the reaction is product favored. For this reaction, G o is less zero at at 291 K. Therefore, the reaction is product favored under standard conditions at this temperature. The change in the Gibbs Free Energy is related to the change in the enthalpy and the entropy. This is summarized by the Gibbs-Helmholtz equation: G o rxn = H o rxn - T S o rxn Rearrange this expression to solve for S o S o = ( H o - G o )/T For the reaction as written at 291 K: S o = -196.0 kJ - -232.6 kJ = 0.1258 kJ 1000 J = 125.8 J/K 291 K K kJ For the reaction of 2.22 moles of H 2 O 2 (l):

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
S = 2.22 mol H 2 O 2 (l) 125.8 J 1 = 140 J/K K 2 mol H 2 O 2 (l) Note that S is positive. The reaction represents one in which the system becomes more RANDOM . For the reaction 2SO 2 (g) + O 2 (g) 2SO 3 (g) G o = -143.5 kJ and S o = -187.9 J/K at 289 K and 1 atm. This reaction is (reactant,product) product favored under standard conditions at 289 K. The standard enthalpy change for the reaction of 2.22 moles of SO 2 (g) at this temperature would be -220 kJ. Feedback: For this reaction, G o is less zero at at 289 K. Therefore, the reaction is product favored under standard conditions at this temperature. The change in the Gibbs Free Energy is related to the change in the enthalpy and the entropy. This is summarized by the Gibbs-Helmholtz equation: G o rxn = H o rxn - T S o rxn Rearrange this expression to solve for H o : H o = G o + T S o For the reaction as written at 289 K: H o = -143.5 kJ + 289 K -187.9 J 1 kJ = -197.8 kJ K 1000 J For the reaction of 2.22 moles of SO 2 (g): H = 2.22 mol SO 2 (g)
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 09/22/2010 for the course CHEM 3 taught by Professor Pedersen during the Fall '08 term at University of California, Berkeley.

### Page1 / 8

Chem HW 9 - Rearrange the equation above to solve for ln...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online