Calculate Delta H r x n for the following reaction

Calculate Delta H r x n for the following reaction: Fe2O3 (s) + 3CO (g)—–> 2Fe (s) + 3CO2 (g). Use the following reactions and given delta H’s. 2Fe (s) + 3/2O2 (g)—–>Fe2O3 (s), Delta H= -824.2 kJ CO (g) + 1/2O2 (g)—–>CO2 (g), Delta H=-282.7 K j

The Correct Answer and Explanation is :

To determine the enthalpy change ((\Delta H_{rxn})) for the reaction:

[
\text{Fe}_2\text{O}_3 (s) + 3CO (g) \rightarrow 2Fe (s) + 3CO_2 (g)
]

using the given reactions:

  1. Formation of Iron(III) Oxide
    [
    2Fe (s) + \frac{3}{2}O_2 (g) \rightarrow Fe_2O_3 (s) \quad \Delta H = -824.2 \text{ kJ}
    ]
  2. Formation of Carbon Dioxide
    [
    CO (g) + \frac{1}{2}O_2 (g) \rightarrow CO_2 (g) \quad \Delta H = -282.7 \text{ kJ}
    ]

Step 1: Reverse the First Equation

The target reaction has Fe(_2)O(_3) as a reactant, but in the given equation, it appears as a product. So, we reverse the equation:

[
Fe_2O_3 (s) \rightarrow 2Fe (s) + \frac{3}{2}O_2 (g)
]

Since we reversed the reaction, we also reverse the sign of (\Delta H):

[
\Delta H = +824.2 \text{ kJ}
]

Step 2: Multiply the Second Equation

The given second reaction forms CO(_2) from CO. Since the target equation involves 3 CO molecules, we multiply the entire equation by 3:

[
3CO (g) + \frac{3}{2}O_2 (g) \rightarrow 3CO_2 (g)
]

Since we multiplied the reaction by 3, we also multiply (\Delta H) by 3:

[
\Delta H = 3 \times (-282.7) = -848.1 \text{ kJ}
]

Step 3: Add the Two Adjusted Equations

Now, summing both reactions:

[
Fe_2O_3 (s) + \frac{3}{2}O_2 (g) + 3CO (g) \rightarrow 2Fe (s) + \frac{3}{2}O_2 (g) + 3CO_2 (g)
]

The (\frac{3}{2}O_2 (g)) cancels from both sides, giving:

[
Fe_2O_3 (s) + 3CO (g) \rightarrow 2Fe (s) + 3CO_2 (g)
]

which matches the target reaction.

Step 4: Sum the Enthalpy Changes

[
\Delta H_{rxn} = 824.2 + (-848.1) = -23.9 \text{ kJ}
]

Conclusion

The enthalpy change for the reaction is (-23.9) kJ, meaning the reaction is exothermic, releasing a small amount of energy. This makes sense because iron oxide is being reduced while carbon monoxide is oxidized to carbon dioxide, a common thermodynamically favorable process in metallurgy.

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