Question

1. Given the values of ΔG_f^o given below in kJ/mol, calculate the value of ΔG^o in kJ for the combustion of 1 mole of methane to form carbon dioxide and gaseous water.
ΔG_f^o (CH₄(g)) = -48
ΔG_f^o (CO₂(g)) = -395
ΔG_f^o (H₂O(g)) = -236

2.

Given the values of S^o given below in J/mol K and of ΔH_f^o given in kJ/mol, calculate the value of ΔG^o in kJ for the combustion of 1 mole of ethane to form carbon dioxide and gaseous water at 298 K.

S (C₂H₆(g)) = 226

S (O₂(g)) = 203

S (CO₂(g)) = 217

S (H₂O(g)) = 181

ΔH_f^o (C₂H₆(g)) = -82

ΔH_f^o (CO₂(g)) = -390

ΔH_f^o (H₂O(g)) = -241

3.

Given the following data,

2 NO(g) + O₂(g) => 2 NO₂(g) ΔG^o = -73

N₂(g) + O₂(g) => 2 NO(g) ΔG^o = 170

2 N₂O(g) => 2 N₂(g) + O₂(g) ΔG^o = -206

Find ΔG^o for N₂O(g) + NO₂(g) => 3 NO(g)

4. At a certain temperature, 110 K, K_p for the reaction,
N₂O₄(g) <=> 2 NO₂(g), is 2.14 x 10⁵⁵.
Calculate the value of DG^o in kJ for the reaction at this temperature.

5. The value of ΔG^o for the precipitation reaction,

Ca²⁺(aq) + CO₃^2-(aq) <=> CaCO₃(s)

is -48.1 kJ at 298 K.

Calculate the value of ΔG in kJ at 298 K if the concentrations of Ca²⁺(aq) and
CO₃^2-(aq) are 0.133 and 0.331, M respectively.

Answer 1

General procedure:

-Write the equation

- identify the species that are in standrd state

- use the appropriate eqution and pay attention to stoichiometric coefficients:

1.

CH4 + 2O2 = CO2 + 2H2O

O2 is in standard state, so dG = 0 . dG = -395 + 2x(-236) – (-48) = -819 kJ/mol

2.

C2H6 + 3.5O2 = 2CO2 + 3H2O

dS = 3x181 + 2x217 – 3.5x203 -226 = …. J/mol.K

dH = …. kJ/mol (similar coefficients)

Convert dS from J/mol.K to kJ/mol.K     (divide the value by 1000)

dG = dH-TdS

3.

(-R1 + 2R2 + R3 ) /2 = R

dG = (73+ 2x170 -206)/2 = ….

4.

dG = - RTlnK

       = - 0.008314 kJ/molKx110K xln2.14x10⁵⁵

       = - 0.914 kJ/mol x 127.4

        = 116.4 kJ/mol

5.

dG = dG^o + RTlnQ         

( reaction quotient Q= 1/ ([CO₃^2-][Ca2+]= 1/(0.133x0.331) = 22.7

dG = -48.1 kJ/mol+ 0.008314kJ/mol.Kx298K x ln 22.7 = …. kJ/mol