Cliffs AP Chemistry, 3rd Edition

Answer: D

This problem requires us to use the Gibbs-Helmholtz equation:

∆G°= ∆H°−T∆S°

Step 1:Using the given information, calculate ∆H°.

∆H°= Σ∆H°products−Σ∆H°reactants

= [2(−100.0)] −[2(−218.0) + 2(−297.0)] = 830.0 kJ/mole

Step 2:Calculate ∆S°.

∆S°= Σ∆S°products−Σ∆S°reactants

= [2(91.0) + 3(205.0)] −[2(70.0) + 2(248.0)]

= 797.0 −636.0 = 161.0 J/mole ⋅K = 0.161 kJ/mole ⋅K

Step 3: Substitute into the Gibbs-Helmholtz equation.

∆G°= ∆H°−T∆S°

. (. ).
mole K

K

1

830 0^298 0 161 782 0

mole

kJ kJ kJ/mole

:

:

-=

3. Given the information that follows, calculate the standard free energy change, ∆G°, for
   the reaction

CH 4 (g) + 2 O 2 (g) →2 H 2 O(,) + CO 2 (g)

Species ∆H°(kJ/mole) at ∆G°(kJ/mole) at

25 °C and 1 atm 25 °C and 1 atm

CH 4 (g) −75.00 −51.00

O 2 (g)0 0

H 2 O(,) −286.00 −237.00

CO 2 (g) −394.00 −394.00

A. −919.00 kJ/mole

B. −817.00 kJ/mole

C. −408.50 kJ/mole

D. 459.50 kJ/mole

E. 919.00 kJ/mole

Part II: Specific Topics