Respuesta :
Answer:
The correct answer is -341.2 kJ per mole.
Explanation:
The reaction given is:
CO (g) + Cl₂ (g) ⇔ COCl₂ (g)
Kp = 5.62 × 10³⁵
T = 25 °C or 298 K
The formula for calculating ΔG is,
ΔG° = -RTlnKp
ΔG° = -8.314 × 298 ln (5.62 × 10^35)
ΔG° = -203.9 kJ/mol
ΔG° = ∑nΔG°f (products) -∑nΔG°f (reactants)
ΔG° = ΔG°f (COCl₂ (g)) - [ΔG°f (CO(g)) + ΔG°f (Cl₂(g))]
ΔG°f (COCl₂ (g)) = ΔG° + [ΔG°f (CO (g)) + ΔG°f (Cl₂(g))]
ΔG°f (COCl₂ (g)) = -203.9 + (-137.28 + 0.00)
ΔG°f (COCl₂ (g)) = -341.2 kJ/mol
The standard Gibbs free energy [tex]\mathbf{\Delta G^o_f}[/tex] for COCl2 at 25°C is -341.25 kJ/mol
The given equation for the chemical reaction is
CO(g) + Cl2(g) ⇌ COCl2(g)
At the temperature of 25°C = (273 + 25) K, the equilibrium constant [tex]\mathbf{K_p = 5.62\times 10^{35}}[/tex]
Consider the expression for the relationship between [tex]\mathbf{\Delta G^o}[/tex] and [tex]\mathbf{K_p }[/tex] for the equilibrium reaction can be expressed as:
[tex]\mathbf{\Delta G^o = - RT In K_p}[/tex]
where;
- gas constant (R) = 8.314 × 10⁻³ kJ/K.mol
∴
[tex]\mathbf{\Delta G^o = - (8.314 \times 10^{-3}\ kJ/K.mol \times 298 \ K) \times In (5.62 \times 10^{35} )}[/tex]
[tex]\mathbf{\Delta G^o = -2.477572\ K \times 82.31680992}[/tex]
[tex]\mathbf{\Delta G^o = 203.95 \ kJ}[/tex]
Thus, the standard free energy for the reaction is 203.95 kJ/mol
For a given reaction, the standard Gibbs free energy can be calculated by using the formula:
[tex]\mathbf{\Delta G^o_{rxn} = \sum n \Delta G^o_f (products) - \sum m \Delta G^o_f (reactants) }[/tex]
[tex]\mathbf{\Delta G^o_{rxn} =\Big [\Delta G^o_{f} (COCl_{2(g)} ) -\Big(\Delta G^o_{f} (CO)_{(g)} + \Delta G^o_{f} (Cl)_{2(g)} ) \Big ) \Big ] }[/tex]
replacing the values of and solving for COCl2 at standard free energy of formation of substances, we have:
[tex]\mathbf{-203.95 \ kJ/mol =\Big [\Delta G^o_{f} (COCl_{2(g)} ) -\Big(-137.3 kJ/mol + 0 \ kJ/mol\Big ) \Big ] }[/tex]
Collecting like terms, we have:
[tex]\mathbf{\Delta G^o_{f} (COCl_{2(g)} ) = -203.95 \ kJ/mol -137.3 kJ/mol }[/tex]
[tex]\mathbf{\Delta G^o_{f} (COCl_{2(g)} ) = -341.25 \ kJ/mol }[/tex]
Therefore, we can conclude that the standard Gibbs free energy [tex]\mathbf{\Delta G^o_f}[/tex] for COCl2 at 25°C is -341.25 kJ/mol
Learn more about standard Gibbs free energy here:
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