Question

1. Given the values of ΔGfo given below in kJ/mol, calculate the value of ΔGo in...

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

2.

Given the values of So given below in J/mol K and of ΔHfo given in kJ/mol, calculate the value of ΔGo 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

ΔHfo (C2H6(g)) = -82

ΔHfo (CO2(g)) = -390

ΔHfo (H2O(g)) = -241

3.

Given the following data,

2 NO(g) + O2(g) => 2 NO2(g) ΔGo = -73

N2(g) + O2(g) => 2 NO(g) ΔGo = 170

2 N2O(g) => 2 N2(g) + O2(g) ΔGo = -206

Find ΔGo for N2O(g) + NO2(g) => 3 NO(g)

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

5. The value of ΔGo for the precipitation reaction,

Ca2+(aq) + CO32-(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
CO32-(aq) are 0.133 and 0.331, M respectively.

Homework Answers

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 = dGo + RTlnQ         

( reaction quotient Q= 1/ ([CO32-][Ca2+]= 1/(0.133x0.331) = 22.7

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

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