Question

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) torches are used in welding. How much heat (in kJ) evolves when 5.0 L of \(C_{2} \mathrm{H}_{2}\) \(\left(d=1.0967 \mathrm{kg} / \mathrm{m}^{3}\right)\) is mixed with a stoichiometric amount of oxygen gas? The combustion reaction is $$\begin{array}{r} \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+\frac{5}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-1299.5 \mathrm{kJ} \end{array}$$

Short answer

The heat evolved when 5.0 L of acetylene is mixed in a stoichiometric amount of oxygen gas is -273.8 kJ

Step-by-step solution

Determine the molecular weight of acetylene

The molecular weight of \(\mathrm{C}_{2} \mathrm{H}_{2}\) is computed by adding the atomic masses of Carbon and Hydrogen. Given that the atomic mass of Carbon (C) is 12 g/mol and that of Hydrogen (H) is 1 g/mol, the molecular weight of \(\mathrm{C}_{2} \mathrm{H}_{2}\) is \(= 2 \times 12 (Carbon) + 2 \times 1 (Hydrogen) = 26 \, \mathrm{g/mol}\)

Determine mass of acetylene

We are given the volume of \(\mathrm{C}_{2} \mathrm{H}_{2}\) to be 5.0 L and its density \(d = 1.0967 \mathrm{kg/m^{3}} = 1.0967 \mathrm{g/cm^{3}}\). Density is \(mass/volume\). Therefore, we calculate mass by multiplying volume by density. So, the mass of acetylene is \(mass = volume \times density = 5.0 L \times 1.0967 \mathrm{g/cm^{3}} = 5.4835 g\)

Find the moles of acetylene

Moles can be calculated using the formula \(moles = mass/molecular \, weight\). So, the moles of acetylene, \(moles_{C2H2} = mass_{C2H2} / molecular \, weight_{C2H2} = 5.4835 g / 26 g/mol = 0.2109 mol\)

Determine heat evolved

The combustion reaction shows that 1 mol of \(\mathrm{C}_{2} \mathrm{H}_{2}\) releases \(-1299.5 kJ\). Therefore, \(0.2109 mol\) of \(\mathrm{C}_{2} \mathrm{H}_{2}\) will release \(∆H = 0.2109 mol \times -1299.5 kJ/mol = -273.8 kJ\)

Key concepts

Enthalpy of Combustion

Enthalpy of combustion describes the amount of heat released during the combustion of a substance with oxygen. In this exercise, acetylene (\(\text{C}_{2}\text{H}_{2}\)) undergoes a combustion reaction. The reaction is represented by \(\text{C}_{2}\text{H}_{2}(g) + \frac{5}{2} \text{O}_{2}(g) \rightarrow 2 \text{CO}_{2}(g) + \text{H}_{2}\text{O}(l) \). The given \(\Delta H^{\circ} = -1299.5 \, \text{kJ}\) indicates the heat change, which is exothermic as shown by the negative sign. This tells us that energy is released into the surroundings.

• Negative \(\Delta H\) values indicate exothermic reactions, where heat is released.
• Each mole of acetylene combusted releases 1299.5 kJ of energy.

Understanding enthalpy in reactions is vital for predicting energy changes and understanding thermodynamics in chemical processes.
It allows us to determine how much energy can be harnessed from a reaction, which is especially important in applications like welding where acetylene torches are used.

Stoichiometry

Stoichiometry involves calculating the quantitative relationships of reactants and products in chemical reactions. It helps determine how substances react stoichiometrically in fixed ratios by moles. In the combustion of acetylene, we see:

• 1 mole of \(\text{C}_{2}\text{H}_{2}\) reacts with \(\frac{5}{2}\) moles of \(\text{O}_{2}\).
• Yields 2 moles of \(\text{CO}_{2}\) and 1 mole of \(\text{H}_{2}\text{O}\).

This stoichiometric ratio is crucial for ensuring optimal combustion with no excess or limiting reagents left. Therefore, knowing how to balance chemical equations and calculate the necessary amounts of materials for a reaction is fundamental.
Stoichiometry ensures reactants in industrial processes are used efficiently and that side products do not form unnecessarily.

Molar Mass

Molar mass is the mass of one mole of a substance, often expressed in grams per mole \(\left(\text{g/mol}\right)\). It is an essential concept for converting between mass and moles in chemical equations. To calculate the molar mass of acetylene (\(\text{C}_{2}\text{H}_{2}\)), add together the atomic masses of carbon and hydrogen:

• Carbon: \(12 \, \text{g/mol}\)
• Hydrogen: \(1 \, \text{g/mol}\)

Thus:\[\text{Molar Mass of } \text{C}_{2}\text{H}_{2} = 2 \times 12 + 2 \times 1 = 26 \text{ g/mol}\]Understanding molar mass is critical in stoichiometry since it allows conversion from grams to moles, facilitating the use of balanced chemical equations to determine specific quantities of reactants or products needed in a reaction.

Density Calculations

Density calculations relate to finding the mass from a given volume or vice versa. Density is expressed as mass per unit volume \(\left(\text{g/cm}^{3}\right)\) or \(\left(\text{kg/m}^{3}\right)\). For the acetylene problem:

• Density of \(\text{C}_{2}\text{H}_{2}\) is \(1.0967 \, \text{g/cm}^{3}\)
• Volume given is \(5.0 \, \text{L}\)

To find mass, use the formula:\[\text{mass} = \text{density} \times \text{volume}\]Hence:\[\text{mass} = 1.0967 \, \text{g/cm}^{3} \times 5.0 \, \text{L}\]Make sure to convert units appropriately when dealing with different volume and density measurements. In this chemical context, accurate density calculations ensure the right amount of material is used for the desired outcome without wastage.