Answered

Solve for \( x \):

[tex]\[ 3x = 6x - 2 \][/tex]



Format the following question or task so that it is easier to read.
Fix any grammar or spelling errors.
Remove phrases that are not part of the question.
Do not remove or change LaTeX formatting.
Do not change or remove [tex] [/tex] tags.
If the question is nonsense, rewrite it so that it makes sense.
-----
Butane \( C_4H_{10}(g) \left( \Delta H_f = -125.6 \, \text{kJ/mol} \right) \) combusts in the presence of oxygen to form \( CO_2(g) \left( \Delta H_f = -393.5 \, \text{kJ/mol} \right) \) and \( H_2O(g) \left( \Delta H_f = -241.82 \, \text{kJ/mol} \right) \) in the reaction:

[tex]\[ 2C_4H_{10}(g) + 13O_2(g) \rightarrow 8CO_2(g) + 10H_2O(g) \][/tex]

What is the enthalpy of combustion, per mole, of butane?

Use [tex]\[ \Delta H_{reaction} = \sum \Delta H_{f, \text{products}} - \sum \Delta H_{f, \text{reactants}} \][/tex]

A. \(-5,315 \, \text{kJ/mol}\)

B. \(-2,657.5 \, \text{kJ/mol}\)

C. \(2,657.4 \, \text{kJ/mol}\)

D. [tex]\(5,314.8 \, \text{kJ/mol}\)[/tex]



Answer :

GinnyAnswer
Let's break down the problem step by step to determine the enthalpy of combustion per mole of butane ([tex]$C_4 H_{10}$[/tex]):

### 1. Write the Balanced Chemical Equation:
[tex]\[ 2 C_4 H_{10} (g) + 13 O_2 (g) \rightarrow 8 CO_2 (g) + 10 H_2O (g) \][/tex]

### 2. Identify the Enthalpy of Formation ([tex]$\Delta H_f$[/tex]):
- [tex]$\Delta H_f$[/tex] of [tex]$CO_2 (g) = -393.5$[/tex] kJ/mol
- [tex]$\Delta H_f$[/tex] of [tex]$H_2O (g) = -241.82$[/tex] kJ/mol
- [tex]$\Delta H_f$[/tex] of [tex]$C_4 H_{10} (g) = -125.6$[/tex] kJ/mol
- [tex]$\Delta H_f$[/tex] for [tex]$O_2 (g)$[/tex] is 0 kJ/mol (since it's in its standard state)

### 3. Calculate the Total Enthalpy of Formation of Reactants:
[tex]\[ \text{Reactants} = (2 \times \Delta H_f \text{ of } C_4 H_{10}) + (13 \times \Delta H_f \text{ of } O_2) \][/tex]
[tex]\[ \text{Reactants} = (2 \times -125.6) + (13 \times 0) = -251.2 \text{ kJ} \][/tex]

### 4. Calculate the Total Enthalpy of Formation of Products:
[tex]\[ \text{Products} = (8 \times \Delta H_f \text{ of } CO_2) + (10 \times \Delta H_f \text{ of } H_2O) \][/tex]
[tex]\[ \text{Products} = (8 \times -393.5) + (10 \times -241.82) = -3148 + (-2418.2) = -5566.2 \text{ kJ} \][/tex]

### 5. Calculate the Enthalpy Change of the Reaction ([tex]$\Delta H_{\text{reaction}}$[/tex]):
[tex]\[ \Delta H_{\text{reaction}} = \sum (\Delta H_f \text{ of products}) - \sum (\Delta H_f \text{ of reactants}) \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -5566.2 - (-251.2) = -5566.2 + 251.2 = -5315.0 \text{ kJ} \][/tex]

### 6. Calculate the Enthalpy of Combustion per Mole of Butane:
Since the given balanced equation represents the combustion of 2 moles of butane, we need to divide the total enthalpy change by 2 to find the enthalpy change per mole of butane.
[tex]\[ \Delta H_{\text{combustion per mole}} = \frac{-5315.0}{2} = -2657.5 \text{ kJ/mol} \][/tex]

### Conclusion:
The enthalpy of combustion per mole of butane is \(-2657.5 \text{ kJ/mol}\).

Therefore, the correct answer is:
[tex]\[ \boxed{-2657.5 \text{ kJ/mol}} \][/tex]

Other Questions