Question
Calculate the standard Gibbs energy of the given reaction at 25°C
CO(g) + half CO +CO2
CO(g) + half CO +CO2
Answers
It is unclear what the products of this reaction are, as the expression provided is incomplete. Please double-check the given chemical equation and provide the full reaction for an accurate calculation of the standard Gibbs energy.
Calculate the standard Gibbs energy of the given reaction at 25°C
CO(g) + half CO(g) =CO2
CO(g) + half CO(g) =CO2
The chemical equation is not balanced, but assuming the balanced equation is:
CO(g) + 1/2 O2(g) → CO2(g)
The standard Gibbs energy change (ΔG°) can be calculated using the following equation:
ΔG° = ΣnΔG°f(products) - ΣnΔG°f(reactants)
where Σn is the sum of the stoichiometric coefficients, and ΔG°f is the standard Gibbs energy of formation for each species at 25°C and 1 atm.
Using standard Gibbs energy of formation values from a table, we have:
ΔG°f(CO2) = -394.36 kJ/mol
ΔG°f(CO) = -137.27 kJ/mol
ΔG°f(O2) = 0 kJ/mol
Substituting these values into the equation and solving, we get:
ΔG° = (1 × -394.36 kJ/mol) - [1 × -137.27 kJ/mol + 1/2 × 0 kJ/mol]
ΔG° = -394.36 kJ/mol + 137.27 kJ/mol
ΔG° = -257.09 kJ/mol
Therefore, the standard Gibbs energy change for this reaction at 25°C is -257.09 kJ/mol.
CO(g) + 1/2 O2(g) → CO2(g)
The standard Gibbs energy change (ΔG°) can be calculated using the following equation:
ΔG° = ΣnΔG°f(products) - ΣnΔG°f(reactants)
where Σn is the sum of the stoichiometric coefficients, and ΔG°f is the standard Gibbs energy of formation for each species at 25°C and 1 atm.
Using standard Gibbs energy of formation values from a table, we have:
ΔG°f(CO2) = -394.36 kJ/mol
ΔG°f(CO) = -137.27 kJ/mol
ΔG°f(O2) = 0 kJ/mol
Substituting these values into the equation and solving, we get:
ΔG° = (1 × -394.36 kJ/mol) - [1 × -137.27 kJ/mol + 1/2 × 0 kJ/mol]
ΔG° = -394.36 kJ/mol + 137.27 kJ/mol
ΔG° = -257.09 kJ/mol
Therefore, the standard Gibbs energy change for this reaction at 25°C is -257.09 kJ/mol.
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