Problem 75 - Chemistry The Molecular Science

Question

Calculate the standard reaction enthalpy, \(\Delta_{\mathrm{t}} H^{\circ}\), for formation of \(1 \mathrm{~mol}\) strontium carbonate (the material that gives the red color in fireworks) from its elements. $$ \mathrm{Sr}(\mathrm{s})+\mathrm{C}(\text { graphite })+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Sr} \mathrm{CO}_{3}(\mathrm{~s}) $$ The information available is $$ \begin{array}{ll} \mathrm{Sr}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SrO}(\mathrm{s}) & \Delta_{\mathrm{r}} H^{\circ}=-592 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{SrO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SrCO}_{3}(\mathrm{~s}) & \Delta_{\mathrm{r}} H^{\circ}=-234 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C}(\text { graphite })+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g}) & \Delta_{\mathrm{r}} H^{\circ}=-394 \mathrm{~kJ} / \mathrm{mol} \end{array} $$

Step-by-Step Solution

Answer

The standard reaction enthalpy, \( \Delta_{\mathrm{t}} H^{\circ} \), is -1220 kJ/mol.

Step 1: Identify Target Reaction

The target reaction is the formation of \( \text{SrCO}_3(\text{s}) \) from \( \text{Sr}(\text{s}), \text{C}(\text{graphite}), \) and \( \text{O}_2(\text{g}) \). We need to determine \( \Delta_{\text{t}} H^{\circ} \) for this reaction: \( \text{Sr}(\text{s}) + \text{C}(\text{graphite}) + \frac{3}{2} \text{O}_2(\text{g}) \rightarrow \text{SrCO}_3(\text{s}) \).

Step 2: Analyze Given Reactions

We have three given reactions with their enthalpy changes:1. \( \text{Sr}(\text{s}) + \frac{1}{2} \text{O}_2(\text{g}) \rightarrow \text{SrO}(\text{s}) \), \( \Delta_{\text{r}} H^{\circ} = -592 \text{ kJ/mol} \)2. \( \text{SrO}(\text{s}) + \text{CO}_2(\text{g}) \rightarrow \text{SrCO}_3(\text{s}) \), \( \Delta_{\text{r}} H^{\circ} = -234 \text{ kJ/mol} \)3. \( \text{C}(\text{graphite}) + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) \), \( \Delta_{\text{r}} H^{\circ} = -394 \text{ kJ/mol} \).

Step 3: Rearrange to Form Target Reaction

To find the enthalpy change for the target reaction, combine the given reactions:1. \( \text{Sr}(\text{s}) + \frac{1}{2} \text{O}_2(\text{g}) \rightarrow \text{SrO}(\text{s}) \)2. \( \text{C}(\text{graphite}) + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) \)3. \( \text{SrO}(\text{s}) + \text{CO}_2(\text{g}) \rightarrow \text{SrCO}_3(\text{s}) \)Add the reactions, ensuring the intermediates \( \text{SrO}(\text{s}) \) and \( \text{CO}_2(\text{g}) \) cancel out, resulting in:\[ \text{Sr}(\text{s}) + \text{C}(\text{graphite}) + \frac{3}{2} \text{O}_2(\text{g}) \rightarrow \text{SrCO}_3(\text{s}) \].

Step 4: Calculate Total Enthalpy Change

Add the enthalpy changes of the reactions to find \( \Delta_{\text{t}} H^{\circ} \):\[ \Delta_{\text{t}} H^{\circ} = (-592 \text{ kJ/mol}) + (-394 \text{ kJ/mol}) + (-234 \text{ kJ/mol}) \]\[ \Delta_{\text{t}} H^{\circ} = -1220 \text{ kJ/mol} \].

Key Concepts

Hess's Law

Hess's Law is a fundamental principle in the study of thermochemistry. It states that the total enthalpy change for a chemical reaction is the same, no matter the route taken, provided the initial and final conditions are identical. This means we can calculate the enthalpy change for a complex reaction by breaking it down into a series of simpler steps, each with a known enthalpy change. By doing this, we follow a pathway of reactions that eventually sum up to the overall desired reaction.

In our exercise, we applied Hess's Law by combining three given reactions to calculate the standard reaction enthalpy for forming strontium carbonate. Checkpoints such as ensuring all intermediate species cancel out, as they do not appear in the final balanced equation, are critical. This confirms that the enthalpy changes add correctly to form the enthalpy change of the target reaction as demanded by Hess's Law.

• This approach allows for flexibility, whereby we can use data from literature or experimental measurements for intermediate steps to determine enthalpies that are otherwise difficult to measure directly.

As long as the same states and pressures are maintained, this technique is reliable for calculating the thermodynamic properties of many reactions.

Standard Reaction Enthalpy

The standard reaction enthalpy, denoted as \( \Delta_{\text{t}} H^{\circ} \), is a measure of the heat change when a reaction occurs under standard conditions, typically 25°C (298 K) and 1 atm pressure.

This quantity provides essential insights into the energetics of chemical reactions. It helps understand whether a reaction will release energy (exothermic) or absorb energy (endothermic). In the case of forming strontium carbonate, the calculated reaction enthalpy is -1220 kJ/mol. This negative sign indicates that the reaction is exothermic, meaning it releases heat to the surroundings.

• Standard enthalpy values are tabulated for standard states of many substances, allowing for quick and accurate calculations using Hess's Law.
• These values are not just academic; they widely aid in practical applications such as designing industrial processes where control over energy transfer is crucial.

Understanding and calculating these values enable chemists to predict the feasibility and energy requirements of reactions they might wish to carry out.

Thermochemistry

Thermochemistry is the branch of chemistry concerned with the quantity of heat evolved or absorbed during chemical reactions. It plays a critical role in both academic and industrial chemistry.

Understanding thermochemistry involves mastering various concepts, including enthalpy, entropy, and free energy. Knowledge of these terms helps chemists identify how energy interplays with chemical transformations. In our exercise, thermochemistry principles guided us in calculating the enthalpy change for forming strontium carbonate, a crucial component of fireworks that gives the characteristic red color when combusted.

• Applying thermochemistry laws like Hess's Law allows chemists to comprehend and quantify energy changes in reactions.
• It's essential for developing safer energy storage, optimizing reaction conditions, and ultimately, enhancing chemical manufacturing efficiency.

For students and practitioners, fluency in thermochemistry is fundamental for exploring and applying chemistry's energy aspects to real-world problems.