Chemistry
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ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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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](https://content.bartleby.com/qna-images/question/ac5cb3b0-7591-4200-b69b-d41956f36b92/1dc82b73-279e-41f6-a11f-bff6278571c3/7cz0vve_processed.jpeg)
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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