Respuesta :
Answer: The [tex]\Delta G[/tex] for the reaction is 55.328 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}=0.100atm\\p_{CCl_4}=0.180atm[/tex]
Putting values in above equation, we get:
[tex]K_p=\frac{(0.735)^2}{0.100\times 0.180}\\\\K_p=30.0125[/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 = 30.0125
Putting values in above equation, we get:
[tex]\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(30.0125))\\\\\Delta G=55327.74J/mol=55.328kJ/mol[/tex]
Hence, the [tex]\Delta G[/tex] for the reaction is 55.328 kJ/mol
The Gibbs free energy of a system gives the spontaneity and the non-spontaneity of the reaction by the amount of the heat released and absorbed. The [tex]\rm \Delta G[/tex] for the reaction is 55.328 kJ/mol.
What is the relationship between the Gibbs free energy and the equilibrium constant?
The balanced chemical reaction can be shown as,
[tex]\rm CO_{2}+CCl_{4} \rightleftharpoons2COCl_{2}[/tex]
Given,
Gibbs free energy of [tex]\rm CO_{2}[/tex] = -394.4 kJ/mol
Gibbs free energy of [tex]\rm CCl_{4}[/tex] = -62.3 kJ/mol
Gibbs free energy of [tex]\rm COCl_{2}[/tex] = -204. 9 kJ/mol
Calculate the [tex]\rm (\Delta G^{\circ}_{rxn})[/tex] Gibbs free energy for the reaction as:
[tex]\rm \Delta G^{\circ}_{rxn} = \sum[n\times \Delta G_{f}(\text{product})] - \sum[n\times \Delta G_{f}(\text{reactant})][/tex]
Substituting values in the equation above:
[tex]\begin{aligned}\rm \Delta G^{\circ}_{rxn} &= \sum[(2\times (-204. 9 ))-((1 \times (-394.4)) + (1\times (-62.3)))]\\\\&= 46900\;\rm J\end{aligned}[/tex]
The equilibrium constant [tex](\rm K_{p})[/tex] is shown as:
[tex]\rm K_{p} =\dfrac {(p_{COCl_{2}})^{2}}{p_{CO_{2}} \times p_{CCl_{4}}}[/tex]
Given,
Pressure of [tex]\rm COCl_{2}[/tex] = 0.735 atm
Pressure of [tex]\rm CO_{2}[/tex] = 0.100 atm
Pressure of [tex]\rm CCl_{4}[/tex] = 0.180 atm
Substituting values in the above equation:
[tex]\begin{aligned}\rm K_{p} &=\dfrac{(0.735)^{2}}{0.100 \times 0.180}\\\\&= 30.01\end{aligned}[/tex]
Calculate the Gibbs free energy from the equilibrium constant by the formula:
[tex]\rm \Delta G = \Delta G^{\circ} + RT \; ln K_{p}[/tex]
Where,
Standard Gibbs's free energy [tex]\rm (\Delta G^{\circ} )[/tex] = 46900 J
Gas constant (R) = 8.314 J/K mol
Temperature (T) = 298 K
Equilibrium constant [tex]\rm (K_{p})[/tex] = 30.01
Substitute the values in the above equation:
[tex]\begin{aligned}\rm \Delta G &= 46900 + (8.314 \times 298 \;\rm ln (30.01))\\\\&= 55327.74 \;\rm J/mol\\\\&= 55.32 \;\rm kJ/mol\end{aligned}[/tex]
Therefore, [tex]\rm \Delta G[/tex] for the reaction is 55.328 kJ/mol.
Learn more about Gibbs free energy and the equilibrium constant here:
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