Respuesta :
Answer: The [tex]\Delta G[/tex] for the reaction is 54.425 kJ/mol
Explanation:
For the given balanced chemical equation:
[tex]CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)[/tex]
We are given:
[tex]\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol[/tex]
To calculate [tex]\Delta G^o_{rxn}[/tex] for the reaction, we use the equation:
[tex]\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)][/tex]
For the given equation:
[tex]\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})][/tex]
Putting values in above equation, we get:
[tex]\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J[/tex]
Conversion factor used = 1 kJ = 1000 J
The expression of [tex]K_p[/tex] for the given reaction:
[tex]K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}[/tex]
We are given:
[tex]p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm[/tex]
Putting values in above equation, we get:
[tex]K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85[/tex]
To calculate the gibbs free energy of the reaction, we use the equation:
[tex]\Delta G=\Delta G^o+RT\ln K_p[/tex]
where,
[tex]\Delta G[/tex] = Gibbs' free energy of the reaction = ?
[tex]\Delta G^o[/tex] = Standard gibbs' free energy change of the reaction = 46900 J
R = Gas constant = [tex]8.314J/K mol[/tex]
T = Temperature = [tex]25^oC=[25+273]K=298K[/tex]
[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = 20.85
Putting values in above equation, we get:
[tex]\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol[/tex]
Hence, the [tex]\Delta G[/tex] for the reaction is 54.425 kJ/mol
The change in free energy for the reaction is 54.4 kJ/mol.
What is change in free energy?
The change in free energy under nonstandard conditions can be obtained from the chage in free energy under standard conditions using the formula;
ΔG = ΔG° + RTlnK
Now ΔG° is obtained from;
ΔG°reaction = 2[−204.9kJ/mol] - [(−394.4kJ/mol) + (−62.3kJ/mol)]
ΔG°reaction =(-409.8) + 456.7
ΔG°reaction = 46.9 kJ/mol
Q =PCOCl2^2/ PCO2 * PCCl4
Q = (0.735)^2/ 0.140 * 0.185
Q = 20.8
ΔG = 46.9 * 10^3 + [8.314 * 298 * ln(20.8)]
ΔG = 54.4 kJ/mol
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