Question

Calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) reacts with water to form acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\). From the following enthalpy of reaction data and data in Appendix C, calculate \(\Delta H_{f}^{\circ}\) for \(\mathrm{CaC}_{2}(s);\) $$ \begin{aligned} \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g) & \\ \Delta H^{\circ}=&-127.2 \mathrm{~kJ} \end{aligned} $$

Short answer

The enthalpy of formation for calcium carbide, \(\mathrm{CaC}_{2}(s)\), is \(-63.5\,\mathrm{kJ/mol}\).

Step-by-step solution

Write down the given reaction and enthalpy change

The given balanced chemical equation for the reaction of calcium carbide with water is: $$ \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g) $$The standard enthalpy change, \(\Delta H^{\circ}\), for this reaction is -127.2 kJ.

Write the general equation for standard enthalpy change

The general equation relating the standard enthalpy change of a reaction to the enthalpies of formation of the species involved is: $$ \Delta H^{\circ} = \sum n_i \Delta H_{f,i}^{\circ}(\text{products}) - \sum n_i \Delta H_{f,i}^{\circ}(\text{reactants}) $$

Find the standard enthalpies of formation for other species

We will find the values of \(\Delta H_{f}^{\circ}\) for the different species involved in the reaction from Appendix C or any standard thermodynamic data table. The values are: $$ \begin{aligned} \Delta H_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O}(l)) &= -285.8\,\mathrm{kJ/mol} \\ \Delta H_{f}^{\circ}(\mathrm{Ca}(\mathrm{OH})_{2}(s)) &= -985.2\,\mathrm{kJ/mol} \\ \Delta H_{f}^{\circ}(\mathrm{C}_{2}\mathrm{H}_{2}(g)) &= 226.7\,\mathrm{kJ/mol} \\ \end{aligned} $$

Substitute the given data in the general equation

Substitute the given values and the values found in Step 3 into the general equation for \(\Delta H^{\circ}\): $$ \begin{aligned} -127.2\,\mathrm{kJ} &= \left[\Delta H_{f}^{\circ}(\mathrm{Ca}(\mathrm{OH})_{2}(s)) + \Delta H_{f}^{\circ}(\mathrm{C}_{2} \mathrm{H}_{2}(g))\right] - \left[\Delta H_{f}^{\circ}(\mathrm{CaC}_{2}(s)) + 2 \Delta H_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O}(l))\right] \end{aligned} $$

Solve for the enthalpy of formation of calcium carbide

Now, rearrange the equation and solve for \(\Delta H_{f}^{\circ}(\mathrm{CaC}_{2})\): $$ \Delta H_{f}^{\circ}(\mathrm{CaC}_{2}) = \Delta H_{f}^{\circ}(\mathrm{Ca}(\mathrm{OH})_{2}(s)) + \Delta H_{f}^{\circ}(\mathrm{C}_{2} \mathrm{H}_{2}(g)) - 2 \Delta H_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O}(l)) + 127.2\,\mathrm{kJ} $$Using the values from Step 3, we have: $$ \Delta H_{f}^{\circ}(\mathrm{CaC}_{2}) = (-985.2\,\mathrm{kJ/mol}) + (226.7\,\mathrm{kJ/mol}) - 2(-285.8\,\mathrm{kJ/mol}) + 127.2\,\mathrm{kJ} $$Calculating the result: $$ \Delta H_{f}^{\circ}(\mathrm{CaC}_{2}) = -63.5\,\mathrm{kJ/mol} $$ So, the enthalpy of formation for calcium carbide, \(\mathrm{CaC}_{2}(s)\), is \(-63.5\,\mathrm{kJ/mol}\).

Key concepts

Calcium Carbide

Calcium carbide, represented as \( \text{CaC}_2 \), is a chemical compound that plays a significant role in the production of acetylene gas. It is commonly used in industries for tasks such as cutting and welding. When calcium carbide reacts with water, it yields acetylene \( \text{C}_2 \text{H}_2 \), which is a valuable fuel, and calcium hydroxide \( \text{Ca(OH)}_2 \). This reaction occurs with the release of heat, showcasing its exothermic nature. Calcium carbide itself appears as a solid and is known for its ability to decompose upon contact with water, making it useful in generating acetylene on-site in various applications. Additionally, its enthalpy of formation, which refers to the energy change when one mole of the compound is formed from its elements in their standard states, provides insight into its stability and reactivity.

Standard Enthalpy Change

The standard enthalpy change \( \Delta H^{\circ} \) of a chemical reaction is the enthalpy change when all reactants and products are in their standard states. It is crucial for understanding how energy is absorbed or released in chemical processes. For the reaction involving calcium carbide and water, the standard enthalpy change is given as \(-127.2 \text{ kJ/mol}\). This negative value indicates that the reaction is exothermic, meaning it releases energy to the surroundings. Understanding the standard enthalpy change helps chemists predict the heat exchange of reactions under standard conditions, which is essential for process design and energy management in industrial applications. Using the general formula for standard enthalpy change, we can determine unknown enthalpies of formation by rearranging the equation to bring the unknown to one side and solving using known values.

Chemical Reactions

Chemical reactions represent the process where substances, known as reactants, transform into different substances, called products. They are usually represented by a chemical equation, which showcases the involved substances and their relationships. In the case of calcium carbide reacting with water, the balanced chemical equation is:

• \( \text{CaC}_2(s) + 2 \text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(s) + \text{C}_2 \text{H}_2(g) \)

This equation indicates that one mole of calcium carbide reacts with two moles of water to produce one mole of calcium hydroxide and one mole of acetylene gas. Balancing chemical equations is crucial because it respects the conservation of mass, ensuring that the quantity of elements remains consistent before and after the reaction. Understanding the stoichiometry of a reaction allows chemists to predict how much product forms and the needed amount of reactants to drive the reaction to completion.