Question

Methane, a component in natural gas, can be used as a fuel in combustion reactions. What is the value for ΔGnon (in kJ) for the following reaction under the given conditions at 293 K? CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g) where ΔHorxn = -803 kJ, ΔSorxn = -4.05 J/K, and [CH4] = 14.51 M, [O2] = 9.27 M, [CO2] = 3.83 M, and [H2O] = 6.41 M. (in kJ)

Answer 1

Answer: The $\Delta G$ for the reaction is -806.86 kJ

Explanation:

We are given:

$\Delta H^o_{rxn}=-803kJ=-803000J$      (Conversion factor:  1 kJ = 1000)

$\Delta S^o_{rxn}=-4.05J/K$

Temperature of the reaction = 293 K

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

$\Delta G^o_{rxn}=\Delat H^o_{rxn}-T\Delta S^o_{rxn}$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=-803000J-[(293K)\times (-4.05J/K)]=-801813.35J$

For the given chemical equation:

$CH_4(g)+2O_2(g)\rightleftharpoons CO_2(g)+2H_2O(g)$

The expression for $K_c$ is given as:

$K_{c}=\frac{[H_2O]^2[CO_2]}{[CH_4][O_2]^2}$

We are given:

$[H_2O]=6.41M$

$[CO_2]=3.83M$

$[CH_4]=14.51M$

$[O_2]=9.27M$

Putting values in above equation, we get:

$K_c=\frac{(6.41)^2\times 3.83}{14.51\times (9.27)^2}$

$K_c=0.126$

To calculate the Gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_c$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = -801813.35 J

R = Gas constant = $8.314J/K mol$

T = Temperature = 293 K

$K_c$ = equilibrium constant in terms of concentration = 0.126

Putting values in above equation, we get:

$\Delta G=-801813.35J+(8.314J/K.mol\times 293K\times \ln(0.126))$

$\Delta G=-806859.46J=-806.86kJ$

Hence, the $\Delta G$ for the reaction is -806.86 kJ