Problem 94 - Chemistry The Science in Context

Question

Baking soda decomposes on heating as follows, creating the holes in baked bread: $$2 \mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)$$ Calculate the standard enthalpy of formation of \(\mathrm{NaHCO}_{3}(s)\) from the following information: $$\begin{aligned} \Delta H_{\mathrm{rxn}}^{\circ} &=-129.3 \mathrm{kJ} & \Delta H_{\mathrm{f}}\left[\mathrm{Na}_{2} \mathrm{CO}_{3}(s)\right] &=-1131 \mathrm{kJ} / \mathrm{mol} \\ \Delta H_{\mathrm{f}}\left[\mathrm{CO}_{2}(g)\right] &=-394 \mathrm{kJ} / \mathrm{mol} & \Delta H_{5}^{\mathrm{C}}\left[\mathrm{H}_{2} \mathrm{O}(\ell)\right] &=-286 \mathrm{kJ} / \mathrm{mol} \end{aligned}$$

Step-by-Step Solution

Answer

Answer: The standard enthalpy of formation of NaHCO3(s) is approximately -1681.7 kJ/mol.

Step 1: Understand the given information

The decomposition reaction for baking soda is given by: $$2 \mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}_{2}\mathrm{CO}_{3}(s) + \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(\ell)$$ We are given the standard enthalpy change of reaction and standard enthalpy of formation for some products and reactants: - \(\Delta H_{\mathrm{rxn}}^{\circ}=-129.3\,\mathrm{kJ}\) - \(\Delta H_{\mathrm{f}}\left[\mathrm{Na}_{2}\mathrm{CO}_{3}(s)\right]=-1131\,\mathrm{kJ} / \mathrm{mol}\) - \(\Delta H_{\mathrm{f}}\left[\mathrm{CO}_{2}(g)\right]=-394\,\mathrm{kJ} / \mathrm{mol}\) - \(\Delta H_{5}^{\mathrm{C}}\left[\mathrm{H}_{2}\mathrm{O}(\ell)\right]=-286\,\mathrm{kJ} / \mathrm{mol}\)

Step 2: Use Hess's Law

According to Hess's Law, the enthalpy change of reaction can be expressed as the sum of enthalpy changes of formation of products minus the sum of the enthalpy changes of the formation of reactants: $$\Delta H_{\mathrm{rxn}}^{\circ} = \sum \Delta H_{\mathrm{f}}^{\circ}(\text{products}) - \sum \Delta H_{\mathrm{f}}^{\circ}(\text{reactants})$$ In our case, we can write the equation as: $$\Delta H_{\mathrm{rxn}}^{\circ} = \left[\Delta H_{\mathrm{f}}\left[\mathrm{Na}_{2}\mathrm{CO}_{3}(s)\right] + \Delta H_{\mathrm{f}}\left[\mathrm{CO}_{2}(g)\right] + \Delta H_{5}^{\mathrm{C}}\left[\mathrm{H}_{2}\mathrm{O}(\ell)\right]\right] - [\Delta H_{\mathrm{f}}(\mathrm{NaHCO}_{3})]$$ We have all the values except for the standard enthalpy of formation of NaHCO3 (\(\Delta H_{\mathrm{f}}(\mathrm{NaHCO}_{3})\)). We'll solve for this value.

Step 3: Rearrange the equation to solve for the unknown enthalpy

Let's isolate the term we want to find, which is the enthalpy of formation of NaHCO3: $$\Delta H_{\mathrm{f}}(\mathrm{NaHCO}_{3}) = \left[\Delta H_{\mathrm{f}}\left[\mathrm{Na}_{2}\mathrm{CO}_{3}(s)\right] + \Delta H_{\mathrm{f}}\left[\mathrm{CO}_{2}(g)\right] + \Delta H_{5}^{\mathrm{C}}\left[\mathrm{H}_{2}\mathrm{O}(\ell)\right]\right] - \Delta H_{\mathrm{rxn}}^{\circ}$$

Step 4: Substitute the known values and calculate the result

Now, we can substitute the given values into the equation and calculate the standard enthalpy of formation of NaHCO3: \(\Delta H_{\mathrm{f}}(\mathrm{NaHCO}_{3}) = (-1131\,\mathrm{kJ/mol} \ +(-394\,\mathrm{kJ/mol}) \ + (-286\,\mathrm{kJ/mol})) - (-129.3\,\mathrm{kJ})\) \(\Delta H_{\mathrm{f}}(\mathrm{NaHCO}_{3}) = -1811\,\mathrm{kJ/mol} + 129.3\,\mathrm{kJ}\) \(\Delta H_{\mathrm{f}}(\mathrm{NaHCO}_{3}) = -1681.7\,\mathrm{kJ/mol}\) The standard enthalpy of formation of NaHCO3(s) is approximately -1681.7 kJ/mol.

Key Concepts

Hess's Law

Hess's Law is an essential principle in chemistry that simplifies the calculation of enthalpy changes in complex reactions. The law states that the total enthalpy change for a chemical reaction is the same, no matter how it is carried out in steps. This means that enthalpy is a state function, and the enthalpy change depends only on the initial and final states, not on the path taken to get from one to the other.

When using Hess's Law, we utilize known enthalpy values to find unknowns by constructing a thermodynamic pathway. We break down complex reactions into simpler steps where enthalpy changes are known. By summing these changes, we obtain the overall enthalpy change for the reaction. This approach is particularly useful when direct measurement of the enthalpy change is challenging.

• For example, suppose we know the enthalpy changes for several reactions that sum up to our target reaction. In that case, we can add and subtract these changes to find the unknown enthalpy.
• Another common use is calculating the standard enthalpy change of reaction based on the standard enthalpies of formation.

Chemical Decomposition

Chemical decomposition is a chemical reaction in which a single compound breaks down into two or more simpler substances. This type of reaction is often represented by the general formula: \[ AB \rightarrow A + B \]Decomposition can occur due to heat, electric current, or other energy sources.

In the example of baking soda (\( \text{NaHCO}_3 \) ) decomposition, heat causes the compound to break down into sodium carbonate (\( \text{Na}_2\text{CO}_3 \)), carbon dioxide (\( \text{CO}_2 \)), and water (\( \text{H}_2\text{O} \)). This reaction helps create the bubbles that make bread rise.

• Decomposition of baking soda is crucial in baking, as it is responsible for the texture of baked goods.
• The process is endothermic, meaning it absorbs heat from the surroundings.

Understanding decomposition helps us grasp how substances interact and transform under different conditions.

Standard Enthalpy Change

The standard enthalpy change of a reaction, denoted as \( \Delta H^\circ_\text{rxn} \), is the heat change associated with a reaction occurring under standard conditions (1 atm pressure and 298.15 K). It reflects how much energy is released or absorbed when reactants convert to products.

Standard enthalpies of formation, \( \Delta H^\circ_f \), are specific enthalpy changes where 1 mole of a substance forms from its elements in their standard states. For example, if calculating the standard enthalpy of formation for baking soda, we align it with known values for sodium carbonate, carbon dioxide, and water.

• When using Hess's Law, the standard enthalpy change of reaction helps determine unknown enthalpies of formation.
• It's an essential tool for predicting reaction energetics and feasibility.

By understanding standard enthalpy changes, students can predict whether reactions are endothermic or exothermic and gauge the energy efficiency of chemical processes. This knowledge has significant applications in laboratory experiments and industrial applications.