EXAMPLE: Consider the reaction CH4(g) + H2O(g)-,3H2(g) CO(g) Using the standard

Question

EXAMPLE:

Consider the reaction CH4(g) + H2O(g)-,3H2(g) CO(g) Using the standard thermodynamic data in the tables linked above, calculate G for this reaction at 298.15K if the pressure of each gas is 19.70 mm Hg ANSWER: 124 kJ/mol Correct The free energy change under non-standard conditions is related to the standard Gibbs Free Energy change for a reaction: Calculate AG°rxn from standard Gibbs free energies of formation linked under "Tables" above Gorxn_ Gf (products) _ Gof (reactants) Moles Species Gof (kJ/rnol) | Products | 3 | H2(g)0.0 Products 3H2(g) 1 CO(g) -137.2 CH4(g) -137.2 Reactants -50.7 H20(g -228.6 0.0 kJ -137.2 kJ 228.6 kJ +1 mol_ -50.7 kJ +1 mol 1 mol 142.1 kJ/mol mol For a gas phase reaction, the thermodynamic equilibrium constant is Kp, so Q must be expressed in terms of partial pressures in atmospheres. The pressure of each gas is then, 1 atm 19.70 mm Hg × 2.592×10-2 atm -2 760 mm Hg Q has the same form as Kp: -2,3 (2.592x10x (2.592x10) 6.719×10 = PCH4(g) x PH20(g) (2.592x102) x (2.592x102) For G in kJ, use R = 0.008314 kJ/mol K: 0.008314 kJ G = G° + RT In Q= 142.1 kJ/mol x 298.15K x In(6.7 1 9810-4) 124.0 kJ/mol + = mol K

Explanation / Answer

(6) CH4(g) + H2O(g) -------> 3H2(g) + CO(g) ; G rxn = ?

G rxn = G0fproducts – G0freactants

G rxn = [(3x G0fH2(g) )+ (G0fCO (g) )]- [(G0fCH4(g) )+ (G0fO2 (g) )]

= [(3x(0))+(-137.2))] - [(-50.7))+ (0)] kJ/mol

         = -86.5 kJ/mol

(7) CO(g) + Cl2(g) -------> COCl2(g) ; G rxn = ?

G rxn = G0fproducts – G0freactants

G rxn = [(G0fCOCl2(g) )]- [(G0fCO(g) )+ (G0fCl2 (g) )]

= [-206.77] - [(-137.2))+ (0)] kJ/mol

         = -69.57 kJ/mol

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