Question

1. using the enthalpies of formation and the data below, determine the efficiency of a butane burning camping stove that is used to heat an aluminum pot of water at a campsite. please find the table tables you will need below this question - you can make it bigger when you need to read the values and then delete it.

a) determine the energy input by calculating the molar enthalpy of combustion (δh°_comb) using standard enthalpies of formation and then determining the energy content of the butane. (hint: 3 steps)

2 c₄h₁₀(g) + 13 o₂(g) → 10 h₂o(l) + 8 co₂(g)

Explanation:

To solve part (a), we follow these steps:

Step 1: Find the mass of butane burned

The mass of butane burned is the initial mass of the burner minus the final mass of the burner.

$$ \text{Mass of } \ce{C4H10} = 32.56\ \text{g} - 16.39\ \text{g} = 16.17\ \text{g} $$

Step 2: Calculate the moles of butane

The molar mass of \(\ce{C4H10}\) is \(4(12.01) + 10(1.008) = 58.12\ \text{g/mol}\).

$$ \text{Moles of } \ce{C4H10} = \frac{16.17\ \text{g}}{58.12\ \text{g/mol}} \approx 0.2782\ \text{mol} $$

Step 3: Use standard enthalpies of formation to find \(\Delta H^\circ_{\text{comb}}\)

The formula for \(\Delta H^\circ_{\text{comb}}\) is:

$$ \Delta H^\circ_{\text{comb}} = \sum n\Delta H^\circ_f(\text{products}) - \sum n\Delta H^\circ_f(\text{reactants}) $$

From standard tables:

• \(\Delta H^\circ_f(\ce{C4H10(g)}) = -124.7\ \text{kJ/mol}\)
• \(\Delta H^\circ_f(\ce{O2(g)}) = 0\ \text{kJ/mol}\) (element in standard state)
• \(\Delta H^\circ_f(\ce{H2O(l)}) = -285.8\ \text{kJ/mol}\)
• \(\Delta H^\circ_f(\ce{CO2(g)}) = -393.5\ \text{kJ/mol}\)

For the reaction \(2\ce{C4H10(g)} + 13\ce{O2(g)}
ightarrow 10\ce{H2O(l)} + 8\ce{CO2(g)}\):

$$
LATEXBLOCK0
$$

$$
LATEXBLOCK1
$$

$$ \Delta H^\circ_{\text{comb (for 2 mol)}} = -6006 - (-249.4) = -5756.6\ \text{kJ} $$

For 1 mol of \(\ce{C4H10}\):

$$ \Delta H^\circ_{\text{comb (per mol)}} = \frac{-5756.6\ \text{kJ}}{2} = -2878.3\ \text{kJ/mol} $$

Step 4: Calculate the energy content of butane

Energy content = moles of butane \(\times\) \(\Delta H^\circ_{\text{comb (per mol)}}\)

$$ \text{Energy input} = 0.2782\ \text{mol} \times (-2878.3\ \text{kJ/mol}) \approx -790\ \text{kJ} $$

(The negative sign indicates exothermic reaction, energy released is 790 kJ)

Final Answer for part (a)

The energy input (energy content of butane burned) is approximately \(\boxed{790\ \text{kJ}}\) (magnitude, as energy released).

Answer:

To solve part (a), we follow these steps:

Step 1: Find the mass of butane burned

The mass of butane burned is the initial mass of the burner minus the final mass of the burner.

$$ \text{Mass of } \ce{C4H10} = 32.56\ \text{g} - 16.39\ \text{g} = 16.17\ \text{g} $$

Step 2: Calculate the moles of butane

The molar mass of \(\ce{C4H10}\) is \(4(12.01) + 10(1.008) = 58.12\ \text{g/mol}\).

$$ \text{Moles of } \ce{C4H10} = \frac{16.17\ \text{g}}{58.12\ \text{g/mol}} \approx 0.2782\ \text{mol} $$

Step 3: Use standard enthalpies of formation to find \(\Delta H^\circ_{\text{comb}}\)

The formula for \(\Delta H^\circ_{\text{comb}}\) is:

$$ \Delta H^\circ_{\text{comb}} = \sum n\Delta H^\circ_f(\text{products}) - \sum n\Delta H^\circ_f(\text{reactants}) $$

From standard tables:

• \(\Delta H^\circ_f(\ce{C4H10(g)}) = -124.7\ \text{kJ/mol}\)
• \(\Delta H^\circ_f(\ce{O2(g)}) = 0\ \text{kJ/mol}\) (element in standard state)
• \(\Delta H^\circ_f(\ce{H2O(l)}) = -285.8\ \text{kJ/mol}\)
• \(\Delta H^\circ_f(\ce{CO2(g)}) = -393.5\ \text{kJ/mol}\)

For the reaction \(2\ce{C4H10(g)} + 13\ce{O2(g)}
ightarrow 10\ce{H2O(l)} + 8\ce{CO2(g)}\):

$$
LATEXBLOCK0
$$

$$
LATEXBLOCK1
$$

$$ \Delta H^\circ_{\text{comb (for 2 mol)}} = -6006 - (-249.4) = -5756.6\ \text{kJ} $$

For 1 mol of \(\ce{C4H10}\):

$$ \Delta H^\circ_{\text{comb (per mol)}} = \frac{-5756.6\ \text{kJ}}{2} = -2878.3\ \text{kJ/mol} $$

Step 4: Calculate the energy content of butane

Energy content = moles of butane \(\times\) \(\Delta H^\circ_{\text{comb (per mol)}}\)

$$ \text{Energy input} = 0.2782\ \text{mol} \times (-2878.3\ \text{kJ/mol}) \approx -790\ \text{kJ} $$

(The negative sign indicates exothermic reaction, energy released is 790 kJ)

Final Answer for part (a)

The energy input (energy content of butane burned) is approximately \(\boxed{790\ \text{kJ}}\) (magnitude, as energy released).