Question

Given the following data $$ \begin{aligned} \mathrm{Ca}(s)+2 \mathrm{C}(\text { graphite }) & \longrightarrow \mathrm{CaC}_{2}(s) & \Delta H &=-62.8 \mathrm{~kJ} \\ \mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CaO}(s) & \Delta H &=-635.5 \mathrm{~kJ} \\ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q) & \Delta H &=-653.1 \mathrm{~kJ} \\ \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-1300 . \mathrm{kJ} \\ \mathrm{C}(\text { graphite })+\mathrm{O}_{2}(g) & \mathrm{CO}_{2}(\mathrm{~g}) & \Delta H &=-393.5 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g) $$

Short answer

The enthalpy change for the reaction \(\mathrm{CaC}_{2}(s) + 2 \mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{Ca(OH)}_{2}(aq) + \mathrm{C}_{2}\mathrm{H}_{2}(g)\) is \(\Delta H = 709.7 \mathrm{kJ}\).

Step-by-step solution

Identify the involved reactions

The given reactions are: (1) \(\mathrm{Ca(s)} + 2 \mathrm{C(graphite)} \longrightarrow \mathrm{CaC}_{2}(s)\); \(\Delta H = -62.8 \mathrm{kJ}\) (2) \(\mathrm{Ca(s)} + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CaO}(s)\); \(\Delta H = -635.5 \mathrm{kJ}\) (3) \(\mathrm{CaO}(s) + \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Ca(OH)}_{2}(aq)\); \(\Delta H = -653.1 \mathrm{kJ}\) (4) \(\mathrm{C}_{2}\mathrm{H}_{2}(g) + \frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l)\); \(\Delta H = -1300 \mathrm{kJ}\) (5) \(\mathrm{C(graphite)} + \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\); \(\Delta H = -393.5 \mathrm{kJ}\)

Manipulate the reactions so they can be added up to the target reaction

The target reaction is: \(\mathrm{CaC}_{2}(s) + 2 \mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{Ca(OH)}_{2}(aq) + \mathrm{C}_{2}\mathrm{H}_{2}(g)\) To obtain the target reaction, we first reverse reaction (1) and add it to the reversed reaction (4). Then, we must add reaction (3) to this summation. Reverse (1): \(\mathrm{CaC}_{2}(s) \longrightarrow \mathrm{Ca(s)} + 2 \mathrm{C(graphite)}\); \(\Delta H = 62.8 \mathrm{kJ}\) Reverse (4): \(2 \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{C}_{2}\mathrm{H}_{2}(g) + \frac{5}{2} \mathrm{O}_{2}(g)\); \(\Delta H = 1300 \mathrm{kJ}\) Add reaction (3): \(\mathrm{CaO}(s) + \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Ca(OH)}_{2}(aq)\); \(\Delta H = -653.1 \mathrm{kJ}\)

Find the enthalpy change for the overall reaction

Now, we add up the reactions and the enthalpy changes: Reverse (1) + Reverse (4) + Reaction (3): \(\mathrm{CaC}_{2}(s) + 2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Ca(OH)}_{2}(aq) + \mathrm{C}_{2}\mathrm{H}_{2}(g)\) \(\Delta H\) for the overall reaction: \(62.8 \mathrm{kJ} + 1300 \mathrm{kJ} - 653.1 \mathrm{kJ} = 709.7 \mathrm{kJ}\) Therefore, the enthalpy change for the given reaction is \(\Delta H = 709.7 \mathrm{kJ}\).

Key concepts

Thermochemistry

Thermochemistry is the branch of chemistry that deals with the study of heat and energy changes that occur during chemical reactions and phase changes. A key concept in thermochemistry is enthalpy (\( H \)), which is a measure of the total energy of a system. When a reaction occurs, the change in enthalpy, denoted as \( \Delta H \), represents the amount of heat absorbed or released under constant pressure.

Understanding enthalpy is crucial because it helps predict whether a reaction will be exothermic (\( \Delta H < 0 \), releasing heat) or endothermic (\( \Delta H > 0 \), absorbing heat). In the context of the given problem, the enthalpy changes for a series of known reactions are provided, and by using these values, we can calculate the \( \Delta H \) of an unknown reaction.

Hess's Law

Hess's Law is a fundamental principle in thermochemistry which states that the total enthalpy change for a chemical reaction is the same, regardless of the number of steps taken from reactants to products. This law makes it possible to calculate the enthalpy change of a complex reaction by breaking it down into a series of simpler steps, whose \( \Delta H \) values are known.

Hess's Law is based on the concept that enthalpy is a state function, meaning it depends only on the initial and final state of the system, and not on the path taken. This principle enables the use of algebraic methods to rearrange and combine known chemical equations to derive the enthalpy change of a target reaction, as demonstrated in the provided solution. Here, the given reactions are manipulated and then added together to find the enthalpy change for the overall chemical process.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In the context of enthalpy changes, each reaction is associated with energy changes in the form of heat. By convention, when a reaction releases heat, it is exothermic, and the \( \Delta H \) is negative. Conversely, when a reaction absorbs heat, it is endothermic, and the \( \Delta H \) is positive.

In the exercise, the calculation of enthalpy change for the target reaction requires understanding both the direction of each reaction and how it relates to the overall equation. By reversing a reaction, we change the sign of \( \Delta H \) to reflect that the energy flow is in the opposite direction. This concept is crucial when applying Hess's Law to calculate the \( \Delta H \) of a new reaction from the sum of several individual reactions with known enthalpy changes.