Answered

Get personalized answers to your unique questions on IDNLearn.com. Our platform is designed to provide quick and accurate answers to any questions you may have.

1. Butane [tex]\(C_4H_{10}(g)\left(\Delta H_f = -125.6 \, \text{kJ/mol}\right)\)[/tex] combusts in the presence of oxygen to form [tex]\(CO_2(g)\left(\Delta H_f = -393.5 \, \text{kJ/mol}\right)\)[/tex] and [tex]\(H_2O(g)\left(\Delta H_f = -241.82 \, \text{kJ/mol}\right)\)[/tex] 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]\(\Delta H_{\text{rxn}} = \sum \left( \Delta H_{\text{f, products}} \right) - \sum \left( \Delta H_{\text{f, reactants}} \right)\)[/tex].

A. [tex]\(-5,315 \, \text{kJ/mol}\)[/tex]
B. [tex]\(-2,657.5 \, \text{kJ/mol}\)[/tex]
C. [tex]\(2,657.4 \, \text{kJ/mol}\)[/tex]
D. [tex]\(5,314.8 \, \text{kJ/mol}\)[/tex]

Mark this and return
Save and Exit
Next
Submit

Sagot :

GinnyAnsweridn
To determine the enthalpy of combustion per mole of butane, we need to follow a few steps.

1. Identify the chemical reaction:
The balanced combustion reaction of butane is:
[tex]\[ 2 \, \text{C}_4\text{H}_{10} (g) + 13 \, \text{O}_2 (g) \rightarrow 8 \, \text{CO}_2 (g) + 10 \, \text{H}_2\text{O} (g) \][/tex]

2. Identify the given enthalpies of formation:
[tex]\[ \Delta H_{f, \text{C}_4\text{H}_{10}} = -125.6 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_{f, \text{CO}_2} = -393.5 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H_{f, \text{H}_2\text{O}} = -241.82 \, \text{kJ/mol} \][/tex]

3. Calculate the sum of enthalpies of formation for the products:
[tex]\[ \sum \Delta H_{f, \text{products}} = 8 \times \Delta H_{f, \text{CO}_2} + 10 \times \Delta H_{f, \text{H}_2\text{O}} \][/tex]
Plugging in the values:
[tex]\[ \sum \Delta H_{f, \text{products}} = 8 \times (-393.5) + 10 \times (-241.82) = -3148 + (-2418.2) = -5566.2 \, \text{kJ} \][/tex]

4. Calculate the sum of enthalpies of formation for the reactants:
[tex]\[ \sum \Delta H_{f, \text{reactants}} = 2 \times \Delta H_{f, \text{C}_4\text{H}_{10}} \][/tex]
Plugging in the values:
[tex]\[ \sum \Delta H_{f, \text{reactants}} = 2 \times (-125.6) = -251.2 \, \text{kJ} \][/tex]

5. Calculate the enthalpy change of the reaction:
[tex]\[ \Delta H_{\text{rxn}} = \sum \Delta H_{f, \text{products}} - \sum \Delta H_{f, \text{reactants}} \][/tex]
Plugging in the values:
[tex]\[ \Delta H_{\text{rxn}} = -5566.2 - (-251.2) = -5566.2 + 251.2 = -5315.0 \, \text{kJ} \][/tex]

6. Determine the enthalpy of combustion per mole of butane:
Since the reaction involves 2 moles of C[tex]\(_4\)[/tex]H[tex]\(_{10}\)[/tex], we need to divide the total enthalpy change by 2:
[tex]\[ \Delta H_{\text{combustion per mole}} = \frac{\Delta H_{\text{rxn}}}{2} = \frac{-5315.0}{2} = -2657.5 \, \text{kJ/mol} \][/tex]

Thus, the enthalpy of combustion per mole of butane is [tex]\(\boxed{-2657.5 \, \text{kJ/mol}}\)[/tex].
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.