Question

The following two reactions are known ${Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g); \Delta H = - 26.8 \, kJ }$ ${FeO(s) + CO (g) -> Fe(s) + CO2(g); \Delta H - 16.5 \, kJ}$ The value of $\Delta H$ for the following reaction ${Fe2O3(s) + CO(g) -> 2FeO(s) + CO2(g) }$ is

• A. - 43.3 kJ
• B. -10.3 kJ
• C. +6.2 kJ
• D. +10.3 kJ

Answer 1

The Correct Option is C

To determine the enthalpy change, $\Delta H$, for the reaction:

${Fe_2O_3(s) + CO(g) \rightarrow 2FeO(s) + CO_2(g)}$

We can use the information from the two given reactions:

1. ${Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g); \ \Delta H = -26.8 \, \text{kJ}}$

2. ${FeO(s) + CO (g) \rightarrow Fe(s) + CO_2(g); \ \Delta H = -16.5 \, \text{kJ}}$

We need to apply Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each individual step. Let's rearrange and add the given reactions to derive the desired reaction:

1. Reverse the second reaction:

   ${Fe(s) + CO_2(g) \rightarrow FeO(s) + CO(g)}; \ \Delta H = +16.5 \, \text{kJ}$

2. Multiply the reversed reaction by 2 to match the coefficients of FeO in the desired reaction:

   ${2Fe(s) + 2CO_2(g) \rightarrow 2FeO(s) + 2CO(g)}; \ \Delta H = 2 \times 16.5 = +33.0 \, \text{kJ}$

3. Add this result to the first reaction:

   ${Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g); \ \Delta H = -26.8 \, \text{kJ}}$

   Combining both:

   The intermediate atoms/molecules (Fe and CO₂) will cancel out, resulting in:

   ${Fe_2O_3(s) + CO(g) \rightarrow 2FeO(s) + CO_2(g)}$

Adding the enthalpy changes of both steps gives:

$\Delta H = (-26.8) + (+33.0) = +6.2 \, \text{kJ}$

Thus, the value of $\Delta H$ for the given reaction is +6.2 kJ.