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Answer the following question. Calculate standard enthalpy of reaction, Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g), from the following data. Δf H°(Fe2O3) = - 824 kJ/mol, - Chemistry

Sum

Answer the following question.

Calculate standard enthalpy of reaction,

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g), from the following data.

Δf H°(Fe2O3) = - 824 kJ/mol,

Δf H°(CO) = - 110 kJ/mol,

Δf H°(CO2) = - 393 kJ/mol

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Solution

Given:

Δf H°(Fe2O3) = - 824 kJ/mol,

Δf H°(CO) = - 110 kJ/mol,

Δf H°(CO2) = - 393 kJ/mol

To find: Standard enthalpy of the given reaction (ΔrH°)

Formula: Δ H° = ∑ Δf H° (products) - ∑ Δf H° (reactants)

Calculation:

The reaction is

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

ΔrH° = ∑ Δf H° (products) - ∑ Δf H° (reactants)

= [2 Δf H° (Fe) + 3 Δf H° (CO2)] - [Δf H° (Fe2O3) + 3 Δf H° (CO)]

= [0 + 3 mol × (- 393 kJ mol-1)] - [1 mol × (- 824 kJ mol-1) + 3 mol × (- 110 kJ mol-1)]

= - 1179 + 824 + 330

= - 25 kJ

The standard enthalpy of the given reaction is –25 kJ.

Concept: Enthalpy (H)
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APPEARS IN

Balbharati Chemistry 12th Standard HSC Maharashtra State Board
Chapter 4 Chemical Thermodynamics
Exercise | Q 4.09 | Page 88
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