Calculate 

Δ*H*orxn

 for the following unbalanced reaction:

→+C2H6*g*  O2*g* + CO2*g*  H2O*g*   

Be sure your answer has the correct number of significant figures.

*Note*: Reference the ***Thermodynamic properties of pure substances*** table for additional information.


To calculate the standard enthalpy change of reaction (\(\Delta H_{\text{rxn}}\)) for the combustion of ethane (\(C_2H_6\)), we'll use the following process:

### Step 1: Write the Balanced Chemical Equation

The given unbalanced reaction is:

\[ C_2H_6 (g) + O_2 (g) \rightarrow CO_2 (g) + H_2O (g) \]

To balance it:

\[ 2C_2H_6 (g) + 7O_2 (g) \rightarrow 4CO_2 (g) + 6H_2O (g) \]

### Step 2: Enthalpy of Formation (\( \Delta H_f^\circ \)) Values

We need to use the standard enthalpies of formation (\(\Delta H_f^\circ\)) from the thermodynamic table. Standard enthalpies of formation for gases are typically:

- \( \Delta H_f^\circ \) of \( C_2H_6 (g) \): \( -84.7 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ \) of \( O_2 (g) \): \( 0 \, \text{kJ/mol} \) (since it's an element in its standard state)
- \( \Delta H_f^\circ \) of \( CO_2 (g) \): \( -393.5 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ \) of \( H_2O (g) \): \( -241.8 \, \text{kJ/mol} \)

### Step 3: Apply the Formula for Enthalpy of Reaction

\[
\Delta H_{\text{rxn}}^\circ = \left( \sum \Delta H_f^\circ (\text{products}) \right) - \left( \sum \Delta H_f^\circ (\text{reactants}) \right)
\]

### Step 4: Calculate

Using the balanced equation, let's calculate for the products and reactants:

\[
\Delta H_{\text{rxn}}^\circ = \left[ 4 \times (-393.5) + 6 \times (-241.8) \right] - \left[ 2 \times (-84.7) + 7 \times (0) \right]
\]

For the products:

\[
4 \times (-393.5) = -1574.0 \, \text{kJ}
\]
\[
6 \times (-241.8) = -1450.8 \, \text{kJ}
\]
\[
\text{Total for products} = -1574.0 + (-1450.8) = -3024.8 \, \text{kJ}
\]

For the reactants:

\[
2 \times (-84.7) = -169.4 \, \text{kJ}
\]
\[
\text{Total for reactants} = -169.4 \, \text{kJ}
\]

### Step 5: Final Calculation

\[
\Delta H_{\text{rxn}}^\circ = -3024.8 \, \text{kJ} - (-169.4 \, \text{kJ}) = -3024.8 + 169.4 = -2855.4 \, \text{kJ}
\]

Thus, the enthalpy change for the reaction (\(\Delta H_{\text{rxn}}\)) is:

\[
\Delta H_{\text{rxn}}^\circ = -2855.4 \, \text{kJ}
\]

This value is rounded to **4 significant figures** as required.

### Conclusion:

The enthalpy change of the reaction (\(\Delta H_{\text{rxn}}\)) is \(-2855 \, \text{kJ}\) (rounded to 4 significant figures).

Would you like more details or explanations about any part of this calculation?

### Related Questions:
1. How do enthalpy changes differ between combustion and other types of reactions?
2. How does the state of water (liquid vs gas) affect the enthalpy of combustion?
3. What role do oxygen's zero enthalpy of formation play in this calculation?
4. How would the reaction enthalpy change if the reaction was incomplete?
5. Can we calculate reaction entropy (\(\Delta S_{\text{rxn}}\)) in a similar way?

### Tip:
Always ensure to balance chemical equations before calculating any thermodynamic properties!

Calculate ΔH°rxn for the following unbalanced reaction: →+C2H6(g) + O2(g) → CO2(g) + H2O(g). Be sure your answer has the correct number of significant figures. Note: Reference the Thermodynamic properties of pure substances table for additional information.

Learn how to calculate the standard enthalpy change (ΔH°rxn) for the combustion of ethane using enthalpy of formation values. This problem involves key concepts in thermochemistry and stoichiometry and applies Hess's Law for accurate reaction enthalpy. Suitable for college-level chemistry students.

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