Consider the balanced reaction 4 A+5 B →3 C+4 D and the enthalpies of formation provided in the table below. AHf Compound (kJ/mol) A 56.4 -399.9 60.7 -273.4 Calculate the AH value for the overall reaction in kJ per mole of reaction. Report your answer to one decimal place (ignore significant figures).

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**Calculating the Enthalpy Change of a Reaction**

Consider the balanced reaction:

\[ 4 \text{A} + 5 \text{B} \rightarrow 3 \text{C} + 4 \text{D} \]

The enthalpies of formation (\(\Delta H_f\)) for the compounds involved in the reaction are provided in the table below:

| Compound | \(\Delta H_f\) (kJ/mol) |
|----------|--------------------------|
| A        | 56.4                     |
| B        | -399.9                   |
| C        | 60.7                     |
| D        | -273.4                   |

**Task:** Calculate the \(\Delta H\) value for the overall reaction in kJ per mole of reaction.

**Instructions:** Report your answer to one decimal place (ignore significant figures).

The enthalpy change (\(\Delta H\)) for the reaction can be calculated using the formula:

\[ \Delta H_{reaction} = \sum \Delta H_f (\text{products}) - \sum \Delta H_f (\text{reactants}) \]

Where:
- \(\sum \Delta H_f (\text{products})\) is the sum of the enthalpies of formation of the products, each multiplied by their respective coefficients from the balanced equation.
- \(\sum \Delta H_f (\text{reactants})\) is the sum of the enthalpies of formation of the reactants, each multiplied by their respective coefficients from the balanced equation.

**Step-by-Step Calculation:**
1. **Calculate the total \(\Delta H_f\) for the products:**
   - \(3 \times \Delta H_f(\text{C}) = 3 \times 60.7 \, \text{kJ/mol} = 182.1 \, \text{kJ/mol}\)
   - \(4 \times \Delta H_f(\text{D}) = 4 \times (-273.4) \, \text{kJ/mol} = -1093.6 \, \text{kJ/mol}\)
   - Total \(\Delta H_f\) for products = \(182.1 \, \text{kJ/mol} + (-1093.6 \, \text{kJ/mol}) = -911.5 \, \text{kJ/mol}\)

2. **Calculate the total \(\Delta
Transcribed Image Text:**Calculating the Enthalpy Change of a Reaction** Consider the balanced reaction: \[ 4 \text{A} + 5 \text{B} \rightarrow 3 \text{C} + 4 \text{D} \] The enthalpies of formation (\(\Delta H_f\)) for the compounds involved in the reaction are provided in the table below: | Compound | \(\Delta H_f\) (kJ/mol) | |----------|--------------------------| | A | 56.4 | | B | -399.9 | | C | 60.7 | | D | -273.4 | **Task:** Calculate the \(\Delta H\) value for the overall reaction in kJ per mole of reaction. **Instructions:** Report your answer to one decimal place (ignore significant figures). The enthalpy change (\(\Delta H\)) for the reaction can be calculated using the formula: \[ \Delta H_{reaction} = \sum \Delta H_f (\text{products}) - \sum \Delta H_f (\text{reactants}) \] Where: - \(\sum \Delta H_f (\text{products})\) is the sum of the enthalpies of formation of the products, each multiplied by their respective coefficients from the balanced equation. - \(\sum \Delta H_f (\text{reactants})\) is the sum of the enthalpies of formation of the reactants, each multiplied by their respective coefficients from the balanced equation. **Step-by-Step Calculation:** 1. **Calculate the total \(\Delta H_f\) for the products:** - \(3 \times \Delta H_f(\text{C}) = 3 \times 60.7 \, \text{kJ/mol} = 182.1 \, \text{kJ/mol}\) - \(4 \times \Delta H_f(\text{D}) = 4 \times (-273.4) \, \text{kJ/mol} = -1093.6 \, \text{kJ/mol}\) - Total \(\Delta H_f\) for products = \(182.1 \, \text{kJ/mol} + (-1093.6 \, \text{kJ/mol}) = -911.5 \, \text{kJ/mol}\) 2. **Calculate the total \(\Delta
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