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ICI(g) + Cl₂(g) → ICl3(s). The AG(f) of ICI(g) is -5.5 kJ/mol and the

AG(f)° of ICl3(s) is -22.59 kJ/mol
What is the AG of the reaction, in kJ/mol?

User Narottam Goyal
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2 Answers

27 votes
27 votes

Final answer:

The ∆G of the reaction ICI(g) + Cl₂(g) → ICl₃(s) is calculated using the standard free energies of formation for the reactants and products, resulting in a ∆G of -17.09 kJ/mol.

Step-by-step explanation:

To find the ∆G of the reaction ICI(g) + Cl₂(g) → ICl₃(s), we need to use the standard free energies of formation (∆G°f) for the reactants and products. The formula for calculating ∆G of a reaction is:

∆G = ∆G°f(products) - ∆G°f(reactants)

According to the question, the ∆G°f for ICI(g) is -5.5 kJ/mol and for ICl₃(s) is -22.59 kJ/mol. Since we have one mole of ICI and ICl₃ each, we can simply plug these values into the formula:

∆G = (-22.59 kJ/mol) - (-5.5 kJ/mol)

∆G = -22.59 kJ/mol + 5.5 kJ/mol

∆G = -17.09 kJ/mol

Therefore, the ∆G of the reaction is -17.09 kJ/mol, indicating that the reaction is spontaneous under standard conditions.

User Tom Stambaugh
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10 votes
10 votes

Answer:

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Step-by-step explanation:

User Legalize
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