Diethyl ether, C 4 H 10 O(l), a flammable compound that was once used as a surgical anesthetic, has the structure The complete combustion of 1 mol of C 4 H 10 O(l) to CO 2 (g) and H 2 O(l) yields ΔH° = -2723.7 kJ. (a) Write a balanced equation for the combustion of 1 mol of C 4 H 10 O(l).

Identify the reactants and products in the combustion reaction. The reactant is diethyl ether, C_4H_10O(l), and the products are carbon dioxide, $$CO_2(g$$), and water, H_2O(l).. Write the unbalanced chemical equation for the combustion of diethyl ether: C_4H_10O(l) + $$O_2(g$$) -\u003e $$CO_2(g) + H_2O(l).. $$Balance the carbon atoms first. There are 4 carbon atoms in C_4H_10O, so you need 4 $$CO_2$$ molecules: C_4H_10O(l) + $$O_2(g$$) -\u003e 4 $$CO_2(g) + H_2O(l).. $$Next, balance the hydrogen atoms. There are 10 hydrogen atoms in C_4H_10O, so you need 5 H_2O molecules: C_4H_10O(l) + $$O_2(g$$) -\u003e 4 $$CO_2(g) + 5 H_2O(l).. $$Finally, balance the oxygen atoms. There are 8 oxygen atoms in 4 $$CO_2$$ and 5 oxygen atoms in 5 H_2O, totaling 13 oxygen atoms needed. Since there is 1 oxygen atom in C_4H_10O, you need 12 more from $$O_2$$, which means 6 $$O_2$$ molecules: C_4H_10O(l) + 6 $$O_2(g$$) -\u003e 4 $$CO_2(g) + 5 H_2O(l).$$

Diethyl ether, C4H10O(l), a flammable compound that was once used as a surgical anesthetic, has the structure The complete combustion of 1 mol of C4H10O(l) to CO2(g) and H2O(l) yields ΔH° = -2723.7 kJ. (a) Write a balanced equation for the combustion of 1 mol of C4H10O(l).

Identify the reactants and products in the combustion reaction. The reactant is diethyl ether, C_4H_10O(l), and the products are carbon dioxide, $$CO_2(g$$), and water, H_2O(l). Write the unbalanced chemical equation for the combustion of diethyl ether: C_4H_10O(l) + $$O_2(g$$) -\u003e $$CO_2(g) + H_2O(l). $$Balance the carbon atoms first. There are 4 carbon atoms in C_4H_10O, so you need 4 $$CO_2$$ molecules: C_4H_10O(l) + $$O_2(g$$) -\u003e 4 $$CO_2(g) + H_2O(l). $$Next, balance the hydrogen atoms. There are 10 hydrogen atoms in C_4H_10O, so you need 5 H_2O molecules: C_4H_10O(l) + $$O_2(g$$) -\u003e 4 $$CO_2(g) + 5 H_2O(l). $$Finally, balance the oxygen atoms. There are 8 oxygen atoms in 4 $$CO_2$$ and 5 oxygen atoms in 5 H_2O, totaling 13 oxygen atoms needed. Since there is 1 oxygen atom in C_4H_10O, you need 12 more from $$O_2$$, which means 6 $$O_2$$ molecules: C_4H_10O(l) + 6 $$O_2(g$$) -\u003e 4 $$CO_2(g) + 5 H_2O(l).$$

Ch.5 - Thermochemistry

Brown - Chemistry: The Central Science 14th Edition

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Ch.5 - Thermochemistry

Problem 78a

Chapter 5, Problem 78a

Diethyl ether, C₄H₁₀O(l), a flammable compound that was once used as a surgical anesthetic, has the structure The complete combustion of 1 mol of C₄H₁₀O(l) to CO₂(g) and H₂O(l) yields ΔH° = -2723.7 kJ. (a) Write a balanced equation for the combustion of 1 mol of C₄H₁₀O(l).

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Identify the reactants and products in the combustion reaction. The reactant is diethyl ether, C_4H_10O(l), and the products are carbon dioxide, CO_2(g), and water, H_2O(l).

Write the unbalanced chemical equation for the combustion of diethyl ether: C_4H_10O(l) + O_2(g) -> CO_2(g) + H_2O(l).

Balance the carbon atoms first. There are 4 carbon atoms in C_4H_10O, so you need 4 CO_2 molecules: C_4H_10O(l) + O_2(g) -> 4 CO_2(g) + H_2O(l).

Next, balance the hydrogen atoms. There are 10 hydrogen atoms in C_4H_10O, so you need 5 H_2O molecules: C_4H_10O(l) + O_2(g) -> 4 CO_2(g) + 5 H_2O(l).

Finally, balance the oxygen atoms. There are 8 oxygen atoms in 4 CO_2 and 5 oxygen atoms in 5 H_2O, totaling 13 oxygen atoms needed. Since there is 1 oxygen atom in C_4H_10O, you need 12 more from O_2, which means 6 O_2 molecules: C_4H_10O(l) + 6 O_2(g) -> 4 CO_2(g) + 5 H_2O(l).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Combustion Reaction

A combustion reaction is a chemical process in which a substance reacts rapidly with oxygen, producing heat and light. In organic compounds, this typically results in the formation of carbon dioxide (CO2) and water (H2O). Understanding the general form of combustion reactions is essential for writing balanced equations, as it helps identify the reactants and products involved.

Balancing Chemical Equations

Balancing chemical equations involves ensuring that the number of atoms of each element is the same on both sides of the equation. This is crucial for obeying the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Techniques such as adjusting coefficients in front of compounds are used to achieve balance.

Enthalpy Change (ΔH°)

Enthalpy change (ΔH°) refers to the heat content change during a chemical reaction at standard conditions. A negative ΔH° indicates that the reaction is exothermic, meaning it releases heat. In the context of combustion, knowing the enthalpy change helps understand the energy released when a compound like diethyl ether is completely combusted, which is important for energy calculations and thermodynamic studies.

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Gasoline is composed primarily of hydrocarbons, including many with eight carbon atoms, called octanes. One of the cleanest–burning octanes is a compound called 2,3,4- trimethylpentane, which has the following structural formula: The complete combustion of one mole of this compound to CO₂(g) and H₂O(g) leads to ΔH° = -5064.9 kJ. (b) By using the information in this problem and data in Table 5.3, calculate H°f for 2,3,4-trimethylpentane.

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Complete combustion of 1 mol of acetone (C₃H₆O) liberates 1790 kJ: C₃H₆O(l) + 4 O₂(g) → 3 CO₂(g) + 3 H₂O(l) ΔH° = -1790 kJ Using this information together with the standard enthalpies of formation of O₂(g), CO₂(g), and H₂O(l) from Appendix C, calculate the standard enthalpy of formation of acetone.