Question

Burning acetylene in oxygen can produce three different carbon-containing products: soot (very fine particles of graphite \(), \mathrm{CO}(g),\) and \(\mathrm{CO}_{2}(g)\). (a) Write three balanced equations for the reaction of acetylene gas with oxygen to produce these three products. In each case assume that \(\mathrm{H}_{2} \mathrm{O}(l)\) is the only other product. (b) Determine the standard enthalpies for the reactions in part (a). (c) Why, when the oxygen supply is adequate, is \(\mathrm{CO}_{2}(g)\) the predominant carbon-containing product of the combustion of acetylene?

Short answer

(a) The balanced equations for the reaction of acetylene with oxygen to produce soot, CO(g), and CO2(g) are: 1. \( C_{2}H_{2}(g) + \frac{5}{2} O_{2}(g) \rightarrow 2C(s) + H_{2}O(l) \) 2. \( 2C_{2}H_{2}(g) + 5O_{2}(g) \rightarrow 4CO(g) + 2H_{2}O(l) \) 3. \( 2C_{2}H_{2}(g) + 5O_{2}(g) \rightarrow 4CO_{2}(g) + 2H_{2}O(l) \) (b) The standard enthalpies for the reactions are: 1. ∆H°(soot) = -797.3 kJ 2. ∆H°(CO) = -1030.8 kJ 3. ∆H°(CO2) = -2602.4 kJ (c) CO2(g) is the predominant carbon-containing product of the combustion of acetylene when the oxygen supply is adequate because its formation is most exothermic. Exothermic reactions are favored because they release energy to the surroundings, making the final state of the system more stable. With adequate oxygen, all the carbon in acetylene is converted to CO2, the most thermodynamically stable form of carbon-containing product.

Step-by-step solution

(a) Writing balanced equations

To write balanced equations, we need to make sure that the number of each type of atom on both sides of the equation is equal. 1. Acetylene (C2H2) reacts with oxygen (O2) to produce soot (graphite, C) and water (H2O). 2. Acetylene (C2H2) reacts with oxygen (O2) to produce carbon monoxide (CO) and water (H2O). 3. Acetylene (C2H2) reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). #Phase 2#

(a) Balancing equations

The balanced equations for each reaction are: 1. \( C_{2}H_{2}(g) + \frac{5}{2} O_{2}(g) \rightarrow 2C(s) + H_{2}O(l) \) 2. \( 2C_{2}H_{2}(g) + 5O_{2}(g) \rightarrow 4CO(g) + 2H_{2}O(l) \) 3. \( 2C_{2}H_{2}(g) + 5O_{2}(g) \rightarrow 4CO_{2}(g) + 2H_{2}O(l) \)

(b) Determining standard enthalpies

We can determine the standard enthalpies (∆H°) of these reactions by using the standard enthalpies of formation (∆H°f) for all reactants and products: ∆H° = Σ [ (moles of product) × (∆H°f of product) ] - Σ [ (moles of reactant) × (∆H°f of reactant) ] The standard enthalpies of formation are as follows: ∆H°f of C2H2(g) = +226.7 kJ/mol ∆H°f of O2(g) = 0 kJ/mol (since it is in its most stable form) ∆H°f of C(s) = 0 kJ/mol (it is the most stable form of carbon) ∆H°f of H2O(l) = -285.8 kJ/mol ∆H°f of CO(g) = -110.5 kJ/mol ∆H°f of CO2(g) = -393.5 kJ/mol Now, we can calculate ∆H° for each reaction: 1. ∆H°(soot) = [ 2 × (-285.8 kJ/mol) ] - [ (+226.7 kJ/mol) + (0 kJ/mol) ] = -797.3 kJ 2. ∆H°(CO) = [ 4 × (-110.5 kJ/mol) + 2 × (-285.8 kJ/mol) ] - [ 2 × (+226.7 kJ/mol) + 5 × (0 kJ/mol) ] = -1030.8 kJ 3. ∆H°(CO2) = [ 4 × (-393.5 kJ/mol) + 2 × (-285.8 kJ/mol) ] - [ 2 × (+226.7 kJ/mol) + 5 × (0 kJ/mol) ] = -2602.4 kJ

(c) Explaining predominant product

When the oxygen supply is adequate, CO2(g) is the predominant carbon-containing product of the combustion of acetylene because its formation is most exothermic, as can be seen from the calculated standard enthalpies of the reactions. Exothermic reactions tend to be favored because the system releases energy to the surroundings, making the final state of the system more stable. In the case of adequate oxygen supply, there is enough oxygen to convert all the carbon in acetylene to CO2, which is the most thermodynamically stable form of carbon-containing product and therefore forms predominantly.

Key concepts

Balanced Chemical Equations

A balanced chemical equation ensures that the number of each type of atom is the same on both sides of the equation. This reflects the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. In the combustion process of acetylene, the steps to balance the equation are meticulous:

• First, identify the reactants: acetylene \( C_2H_2 \) and oxygen \( O_2 \).
• Then, state the possible products: soot (graphite \( C \)), carbon monoxide \( CO \), and carbon dioxide \( CO_2 \) with water \( H_2O \).
• Finally, adjust coefficients to ensure atoms are equal on both sides for each reaction scenario:

- Graphite Production: \( C_2H_2(g) + \frac{5}{2} O_2(g) \rightarrow 2C(s) + H_2O(l) \)
- Carbon Monoxide: \( 2C_2H_2(g) + 5O_2(g) \rightarrow 4CO(g) + 2H_2O(l) \)
- Carbon Dioxide: \( 2C_2H_2(g) + 5O_2(g) \rightarrow 4CO_2(g) + 2H_2O(l) \)
Balancing equations is essential as it also enables effective calculation of other quantities in reactions, such as enthalpy changes.

Standard Enthalpy of Formation

The standard enthalpy of formation \( \Delta H^\circ_f \) is the heat change, under standard conditions, when one mole of a substance is formed from its elements in their most stable forms. It is a valuable measure in calculating reaction enthalpies for combustion reactions. Here's how it aids in the exercise:

• For acetylene combustion, the standard enthalpies of formation of all involved substances are known, except for elements in their standard states (for example, \( O_2 \) and graphite \( C \) are zero).
• The reaction enthalpy \( \Delta H^\circ \) is calculated using the formula:
  \[ \Delta H^\circ = \sum [( \text{moles of products} ) \times ( \Delta H^\circ_f \text{ of products} )] - \sum [( \text{moles of reactants} ) \times ( \Delta H^\circ_f \text{ of reactants} )] \]

This process allows for determination of whether a reaction is exothermic (\( \Delta H^\circ \) is negative) or endothermic (\( \Delta H^\circ \) is positive). For the complete combustion to \( CO_2 \), the enthalpy proves to be highly exothermic, thus releasing significant energy.

Thermodynamic Stability

Thermodynamic stability refers to the stability of a compound as determined by its standard enthalpy of formation. A more negative enthalpy of formation suggests a compound is lower in energy and thus more stable. In combustion reactions, stability is key in determining the preferred products.

• The formation of \( CO_2 \) during acetylene combustion is a prime example. When given enough oxygen, all carbon in acetylene primarily forms \( CO_2 \), as it is the most thermodynamically stable product.
• Exothermic reactions that provide a large negative change in enthalpy (such as the formation of \( CO_2 \) from acetylene) are typically favored, because they decrease the system's energy, thus enhancing stability.
• This preference for \( CO_2 \) over other products like soot or \( CO \) is evident from the substantial enthalpy difference calculated earlier.

Thus, thermodynamic stability guides the outcomes of combustion reactions, favoring products that yield more stable chemical systems.