Problem 91 - Chemistry and Chemical Reactivity

Question

(a) Calculate the enthalpy change, \(\Delta H^{\circ},\) for the formation of 1.00 mol of strontium carbonate (the material that gives the red color in fireworks) from its elements. $$\operatorname{Sr}(\mathrm{s})+\mathrm{C}(\text { graphite })+\frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \operatorname{Sr} \mathrm{CO}_{3}(\mathrm{s})$$ The experimental information available is $$\begin{aligned}\mathrm{Sr}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow & \mathrm{SrO}(\mathrm{s}) & \Delta H_{f}^{\circ} &=-592 \mathrm{kJ} \\\\\mathrm{SrO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \longrightarrow & \mathrm{SrCO}_{3}(\mathrm{s}) & \Delta H_{\mathrm{rxn}}^{\circ} &=-234 \mathrm{kJ} \\\\\mathrm{C}(\mathrm{graphite})+\mathrm{O}_{2}(\mathrm{g}) & \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) & \Delta H_{f}^{0} &=-394 \mathrm{kJ}\end{aligned}$$ (b) Draw an energy level diagram relating the energy quantities in this problem.

Step-by-Step Solution

Answer

\(\Delta H^{\circ} = -432\, \mathrm{kJ/mol}\).

Step 1: Write Down Known Reactions

We have three known reactions with their enthalpy changes:\[ \operatorname{Sr}(\text{s}) + \frac{1}{2} \mathrm{O}_2(\text{g}) \rightarrow \mathrm{SrO}(\text{s}) \quad \Delta H_f^{\circ} = -592\, \mathrm{kJ} \] \[ \mathrm{SrO}(\text{s}) + \mathrm{CO}_2(\text{g}) \rightarrow \mathrm{SrCO}_3(\text{s}) \quad \Delta H_{\mathrm{rxn}}^{\circ} = -234\, \mathrm{kJ} \] \[ \mathrm{C}(\text{graphite}) + \mathrm{O}_2(\text{g}) \rightarrow \mathrm{CO}_2(\text{g}) \quad \Delta H_f^{0} = -394\, \mathrm{kJ} \]

Step 2: Reverse and Adjust the Order of Intermediate Reactions

To derive the target reaction, we need to manipulate these equations. First, consider reversing the third equation to express the decomposition of CO\(_2\) into C and O\(_2\), requiring the enthalpy change to be \( +394\) kJ since the sign is changed: \[ \mathrm{CO}_2(\mathrm{g}) \rightarrow \mathrm{C}(\text{graphite}) + \mathrm{O}_2(\mathrm{g}) \].

Step 3: Combine Adjusted Reactions

Combine the modified reactions to verify if they sum to the overall formation reaction for SrCO\(_3\):\[ \begin{align*} \operatorname{Sr}(\text{s}) + \frac{1}{2} \mathrm{O}_2(\text{g}) &\rightarrow \mathrm{SrO}(\text{s}) &\Delta H = -592\, \mathrm{kJ} \ \mathrm{SrO}(\text{s}) + \mathrm{CO}_2(\text{g}) &\rightarrow \mathrm{SrCO}_3(\text{s}) &\Delta H = -234\, \mathrm{kJ} \ \mathrm{CO}_2(\text{g}) &\rightarrow \mathrm{C}(\text{graphite}) + \mathrm{O}_2(\text{g}) &\Delta H = +394\, \mathrm{kJ} \end{align*}\] These equations sum up to give: \[ \operatorname{Sr}(\text{s}) + \mathrm{C}(\text{graphite}) + \frac{3}{2} \mathrm{O}_2(\text{g}) \rightarrow \operatorname{SrCO}_3(\text{s}) \].

Step 4: Calculate \(\Delta H^{\circ}\) for the Formation

Add the enthalpy changes of the combined reactions to get the net enthalpy change for the formation of SrCO\(_3\): \[ \Delta H^{\circ} = -592\, \mathrm{kJ} + (-234\, \mathrm{kJ}) + 394\, \mathrm{kJ} = -432\, \mathrm{kJ} \]. The \(\Delta H^{\circ}\) of the original reaction is -432 kJ/mol.

Step 5: Draw the Energy Level Diagram

Start with Sr, C, and \(\frac{3}{2}\)O\(_2\) on an energy level, and use arrows to show the stepwise formation of SrO (down with -592 kJ), formation of SrCO\(_3\) from SrO and CO\(_2\) (down with an additional -234 kJ), and decomposition of CO\(_2\) to C and O\(_2\) (up with +394 kJ). The net arrow from the reactants to the product SrCO\(_3\) should point downward, showing the net enthalpy of -432 kJ.

Key Concepts

Formation Reactions

Formation reactions focus on creating a compound from its basic elements in their natural states. Let's take the example of strontium carbonate, SrCO extsubscript{3}, which forms from strontium (Sr), carbon (C as graphite), and oxygen (O extsubscript{2}). These are the basic elements needed for SrCO extsubscript{3}.
A formation reaction is written such that it forms exactly one mole of the compound. This is necessary because the standard enthalpy change of formation, \( \Delta H_f^{\circ} \), represents the energy change when one mole of a compound is formed from its elements under standard conditions (25°C and 1 atm). For SrCO extsubscript{3}, the reaction is:
\[ \mathrm{Sr}( ext{s})+\mathrm{C}( ext{graphite})+\frac{3}{2} \mathrm{O}_{2}( ext{g}) \longrightarrow \mathrm{SrCO}_{3}( ext{s}) \]
Understanding formation reactions is essential as they serve as building blocks in many chemical processes and help calculate important thermodynamic properties like enthalpy changes.

Enthalpy Diagrams

Enthalpy diagrams visually represent the energy changes that occur during a chemical reaction. These diagrams help illustrate the concept of energy absorbed or released. For the SrCO extsubscript{3} formation, an enthalpy diagram can be particularly enlightening.

• Begin with the reactants' energy level: Sr, C, and \( \frac{3}{2} \)O extsubscript{2} at a higher energy level compared to the product.
• Draw a downward arrow to indicate the formation of SrO, releasing 592 kJ of energy.
• Another downward step indicates SrCO extsubscript{3} formation from SrO and CO extsubscript{2} with an additional release of 234 kJ.
• Finally, an upward arrow shows the decomposition of CO extsubscript{2} back to C and O extsubscript{2}, requiring +394 kJ.

This results in an overall downward arrow from reactants to product, showing a net release of energy (\( \Delta H^{\circ} = -432 \) kJ). This diagram helps visualize how the energy transitions lead to the formation of SrCO extsubscript{3}.

Hess's Law

Hess's Law is a crucial principle in thermochemistry, stating that the total enthalpy change for a chemical reaction is the same, regardless of the pathway taken, as long as the initial and final conditions are the same. This property allows us to calculate the enthalpy change of reactions that may be difficult to study directly.
For example, in the formation of 1 mole of SrCO extsubscript{3}, we didn't have the direct enthalpy change. Instead, we used known reactions to construct a pathway.

• We first used the reaction of Sr with O extsubscript{2} to create SrO with \( \Delta H = -592 \) kJ.
• Next, we combined SrO with CO extsubscript{2}, leading to SrCO extsubscript{3}, releasing another 234 kJ.
• Finally, breaking CO extsubscript{2} into C and O extsubscript{2} consumed 394 kJ.

By adding these steps together, using Hess's Law, we calculated the \( \Delta H^{\circ} \) of the overall reaction as -432 kJ/mol. This shows how reactions can be broken down into simpler steps to determine energy changes easily.