Problem 73 - Chemistry An Atoms First Approach

Question

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

Step-by-Step Solution

Answer

The enthalpy change for the reaction \( \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \) is -2006.6 kJ.

Step 1: Identify the target reaction

We are given the following target reaction: $$\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g)$$

Step 2: Rearrange and manipulate the given reactions

We need to manipulate the given reactions to obtain the target reaction as follows: 1. Reverse reaction 1 and multiply it by 1: $$ \mathrm{CaC}_{2}(s) \longrightarrow \mathrm{Ca}(s)+2 \mathrm{C}(\mathrm{graphite}) \quad \quad \Delta H = +62.8 \mathrm{kJ}$$ 2. Reverse reaction 3: $$ \mathrm{Ca}(\mathrm{OH})_{2}(aq) \longrightarrow \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \quad \quad \Delta H = +653.1 \mathrm{kJ}$$ 3. Add reaction 2 to the reversed reaction 3: $$ \mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) + \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CaO}(s)+ \mathrm{Ca}(\mathrm{OH})_{2}(aq) \quad \quad \Delta H = -635.5 + 653.1 \mathrm{kJ}$$ 4. Keep the reaction 4 as is. 5. Multiply reaction 5 by 2 and keep it as it is. We will now add all these new reactions to find the target reaction and its enthalpy change.

Step 3: Add the manipulated reactions

Add all the manipulated reactions together: $$ \mathrm{CaC}_{2}(s) \longrightarrow \mathrm{Ca}(s)+2 \mathrm{C}(\mathrm{graphite}) \quad \quad \Delta H = +62.8 \mathrm{kJ}$$ $$ \mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) + \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CaO}(s)+ \mathrm{Ca}(\mathrm{OH})_{2}(aq) \quad \quad \Delta H = 17.6 \mathrm{kJ}$$ $$\mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \quad \quad \Delta H = -1300 \mathrm{kJ}$$ $$ 2[\mathrm{C}(\text {graphite})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})] \quad \quad \Delta H = -787 \mathrm{kJ}$$ Cancelling out common terms from both sides of the reactions, we obtain the target reaction: $$\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g)$$

Step 4: Calculate the enthalpy change of the target reaction

Now, we simply add the enthalpy changes of all the manipulated reactions to find the enthalpy change of the target reaction: $$\Delta H = +62.8 + 17.6 -1300 -787 \mathrm{kJ}$$ $$\Delta H = -2006.6 \mathrm{kJ}$$ Thus, the enthalpy change for the given reaction is -2006.6 kJ.

Key Concepts

Thermochemistry

Thermochemistry is a branch of chemistry that involves the study of heat and energy changes associated with chemical reactions and physical transformations. The most fundamental concept in thermochemistry is the enthalpy change (ΔH), which is the amount of heat released or absorbed at constant pressure during a chemical reaction.

When a reaction releases heat into the surroundings, it is exothermic, and the enthalpy change is negative. Conversely, when a reaction absorbs heat from the surroundings, it is endothermic, and the enthalpy change is positive.

In the context of the exercise we are discussing, calculating the enthalpy change is crucial. It allows us to quantify the energy that would be released or absorbed if the reaction were to take place under standard conditions. Understanding how to manipulate and interpret enthalpy changes, such as those given in the exercise, is a key skill in thermochemistry.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, during which bonds between atoms are broken and new bonds are formed. This process is governed by the laws of thermodynamics and can involve the transfer of energy between the system and its surroundings.

To solve the given exercise, knowledge of chemical reactions is imperative, particularly regarding the ability to balance equations and understand stoichiometry. Stoichiometry is the quantitative relationship among substances as they participate in chemical reactions. It's essential for predicting the amounts of reactants consumed and products formed in a given reaction.

In addition to balancing the overall equation, considering the states of substances (solid, liquid, gas, aqueous) provides insight into the interaction between the chemicals and the energy changes involved. These factors all play a role in the comprehensive understanding of a chemical process such as the one described in our exercise.

Hess's Law

Hess's Law is a statement in chemistry that states the total enthalpy change during the course of a chemical reaction is the same whether the reaction is made in one step or several steps. This principle is fundamental when we're unable to measure the enthalpy change of a reaction directly or when the reaction does not occur in a single step.

In our exercise, Hess's Law is applied to find the ΔH by manipulating and combining several given chemical reactions to arrive at the target reaction. Key to this approach is understanding that the enthalpy change of a reaction is dependent only on the initial and final states and not on the path taken.

By reversing, multiplying, or dividing the given reactions, we can cancel out intermediate components and construct a pathway that represents the target chemical equation. The sum of the enthalpy changes of these manipulated reactions gives us the overall enthalpy change for the target reaction, demonstrating the practical application of Hess's Law.