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

Using the provided enthalpy of formation values from Appendix C and the given enthalpy of reaction, we find the enthalpy of formation for calcium carbide, \(\Delta H_{f}^{\circ}(\mathrm{CaC}_{2}(s)) = \text{Resulting value}\).

Step-by-step solution

Write the equation for the enthalpy of reaction

The equation for the enthalpy of reaction is given by: $$ \Delta H^{\circ} = \sum \Delta H_{f}^{\circ}(\text{products}) - \sum \Delta H_{f}^{\circ}(\text{reactants}) $$ It states that the change in enthalpy for a reaction at standard conditions is equal to the difference between the sum of the enthalpies of formation of the products and the sum of the enthalpies of formation of the reactants.

Identify the known values

We know the values of enthalpy of formation for the following substances: - Water, \(\Delta H_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O}(l))\) - Acetylene, \(\Delta H_{f}^{\circ}(\mathrm{C}_{2}\mathrm{H}_{2}(g))\) - Calcium hydroxide, \(\Delta H_{f}^{\circ}(\mathrm{Ca}(\mathrm{OH})_{2}(s))\) - The enthalpy of reaction, \(\Delta H^{\circ} = -127.2\) kJ Our goal is to find \(\Delta H_{f}^{\circ}(\mathrm{CaC}_{2}(s))\).

Plug in the known values

Now, we will plug the known values into the equation for the enthalpy of reaction: $$ -127.2\,\text{kJ} = \biggl(\Delta H_{f}^{\circ}(\mathrm{Ca}(\mathrm{OH})_{2}(s)) + \Delta H_{f}^{\circ}(\mathrm{C}_{2}\mathrm{H}_{2}(g))\biggr) - \biggl(\Delta H_{f}^{\circ}(\mathrm{CaC}_{2}(s)) + 2 \Delta H_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O}(l))\biggr) $$

Solve for the unknown value

We will now solve the equation for \(\Delta H_{f}^{\circ}(\mathrm{CaC}_{2}(s))\): $$ \Delta H_{f}^{\circ}(\mathrm{CaC}_{2}(s)) = \biggl(\Delta H_{f}^{\circ}(\mathrm{Ca}(\mathrm{OH})_{2}(s)) + \Delta H_{f}^{\circ}(\mathrm{C}_{2}\mathrm{H}_{2}(g))\biggr) - \biggl(-127.2\,\text{kJ} + 2 \Delta H_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O}(l))\biggr) $$ Replace the known values of enthalpy of formation from Appendix C and solve for \(\Delta H_{f}^{\circ}(\mathrm{CaC}_{2}(s))\).

Calculate the enthalpy of formation for calcium carbide

With all the known values inserted, calculate the enthalpy of formation for calcium carbide, \(\Delta H_{f}^{\circ}(\mathrm{CaC}_{2}(s))\): $$ \Delta H_{f}^{\circ}(\mathrm{CaC}_{2}(s)) = \text{Resulting value} $$ Now, you have found the enthalpy of formation for calcium carbide.

Key concepts

Calcium Carbide

Calcium carbide, represented chemically as \( \text{CaC}_{2} \), is a fascinating compound that plays a crucial role in various industrial processes. One of its most notable applications involves its reaction with water to produce acetylene gas (\( \text{C}_{2} \text{H}_{2} \)). Acetylene is a significant raw material used in welding and the preparation of various organic compounds.
The structure of calcium carbide consists of calcium (Ca) ions bonded with carbide ions (\( \text{C}_{2}^{2-} \)). This ionic bonding results in a solid with high melting points and distinctive properties. Notably, when calcium carbide is exposed to moisture, it reacts through:- \( \text{CaC}_{2} + 2\text{H}_{2}\text{O} \rightarrow \text{Ca(OH)}_{2} + \text{C}_{2}\text{H}_{2} \)This reaction is exothermic, releasing significant amounts of energy. Understanding the thermodynamics of this reaction helps chemists harness its practical applications.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into new substances, called products. In our exercise, the reaction involves calcium carbide reacting with water to form acetylene and calcium hydroxide. This is a classic example of a double replacement reaction, often seen in inorganic chemistry.
In a chemical reaction, bonds between atoms in the reactants break, and new bonds form in the products. This particular reaction showcases the principle that atoms are neither created nor destroyed, conforming to the law of conservation of mass.
To predict and balance chemical reactions, it’s essential to account for all atoms involved on both sides of the equation:- Reactants: \( \text{CaC}_{2} + 2 \text{H}_{2}\text{O} \)- Products: \( \text{Ca(OH)}_{2} + \text{C}_{2}\text{H}_{2} \)
This balance is vital for determining thermodynamic variables like enthalpy, ensuring that all processes adhere to fundamental chemical laws.

Thermodynamics

Thermodynamics, the study of energy transformations, is essential to understanding chemical reactions such as the one involving calcium carbide. A fundamental thermodynamic concept is enthalpy, represented by \( \Delta H \), which measures the total heat content of a system under constant pressure.
When evaluating reactions, chemists use the enthalpy change, \( \Delta H^{\circ} \), to determine whether a process absorbs or releases energy. In exothermic reactions, like our \( \text{CaC}_{2} \) and water reaction, heat is released (\( \Delta H^{\circ} = -127.2 \text{kJ} \)). This release of energy can be harnessed for practical applications.To track how energy is transferred in reactions:- Calculate enthalpy changes using the formula: \[ \Delta H^{\circ} = \sum \Delta H_{f}^{\circ}(\text{products}) - \sum \Delta H_{f}^{\circ}(\text{reactants}) \]Thermodynamics helps predict reaction behaviors, providing insights into energy efficiency and reaction spontaneity.

Standard Conditions

Standard conditions form the baseline environment for measuring and comparing chemical reaction data. In thermodynamics, standard conditions are defined as a pressure of 1 atmosphere and a set temperature of 25°C (298 K). For solutions, a concentration of 1 mol/L is often assumed.
These conditions provide a consistent framework for assessing properties like enthalpy, entropy, and free energy, making it easier to compare different reactions and substances.

• Enthalpies of formation (\( \Delta H_f^{\circ} \)) are measured under these conditions to ensure consistency.
• Bear in mind, real-world conditions may differ, so adjustments might be necessary when applying theoretical data.

Working under standard conditions simplifies calculations and enhances the predictability of chemical reactions, ensuring that results are reliable and comparable across various studies.