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Calculate reaction 7 and 8 for standard enthalpy change of formation.

Reaction 7 calculate for Cr2O3
Reaction 8 calculate for CO2

Calculate reaction 7 and 8 for standard enthalpy change of formation. Reaction 7 calculate-example-1
User Madalena
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The standard enthalpy change of formation for Cr₂O₃(g) is 330 KJ/mol and for CO₂(g) is -393.53 KJ/mol.

We can use Hess's law and the data provided to calculate the standard enthalpy change of formation for Cr₂O₃(g) and CO₂(g).

For reaction 7, we need to find the standard enthalpy change of formation for Cr₂O₃(g). To do this, we can use the given equation and the standard enthalpy changes of formation for CO(g) and Cr(g). Using the equation:

3C(s) + Cr₂O₃(g) → 2Cr(g) + 3CO(g)

and the given ΔHr⁰ values:

ΔHr⁰ for 3CO(g) = 110 KJ/mol

we can calculate the standard enthalpy change of formation for Cr2O3(g) as follows:

ΔHf⁰ for Cr₂O₃(g) = [2(ΔHr⁰ for Cr(g))] + [3(ΔHr⁰ for CO(g))]

ΔHf⁰ for Cr₂O₃(g) = [2(0)] + [3(110)] = 330 KJ/mol

For reaction 8, we need to find the standard enthalpy change of formation for CO₂(g). To do this, we can use the given equation and the standard enthalpy changes of formation for CH₃OH(g) and H₂O(g). Using the equation:

CH₃OH(g) + 1¹/2O₂(g) → CO₂(g) + 2H₂O(g)

and the given ΔHr⁰ values:

ΔHr⁰ for CH₃OH(g) = -239.10 KJ/mol

ΔHr⁰ for H₂O(g) = -241.83 KJ/mol

we can calculate the standard enthalpy change of formation for CO₂(g) as follows:

ΔHf⁰ for CO₂(g) = [ΔHr⁰ for CO₂(g)] - [ΔHr⁰ for CH₃OH(g)] - [2(ΔHr⁰ for H₂O(g)]

ΔHf⁰ for CO₂(g) = [-638.06] - [-239.10] - [2(-241.83)] = -393.53 KJ/mol

User Livingston
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