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?

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#

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. \( 2C_{2}H_{2}(g) + 5O_{2}(g) \rightarrow 4CO_{2}(g) + 2H_{2}O(l) \)

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

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.