Question

Calculate \(\Delta H^{\circ}\) for each of the following reactions using the data in Appendix 4: $$ \begin{array}{c} 4 \mathrm{Na}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Na}_{2} \mathrm{O}(s) \\ 2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g) \\ 2 \mathrm{Na}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{CO}(\mathrm{g}) \end{array} $$ Explain why a water or carbon dioxide fire extinguisher might not be effective in putting out a sodium fire.

Short answer

The standard enthalpy change for each reaction is as follows: 1) \(∆H° = -832\) kJ/mol 2) \(∆H° = -364.6\) kJ/mol 3) \(∆H° = -133\) kJ/mol Water is not suitable for extinguishing sodium fires because it reacts with sodium to produce sodium hydroxide and combustible hydrogen gas. Carbon dioxide is also not effective as it reacts with sodium to form sodium oxide and carbon monoxide, which do not help in extinguishing the fire. A special Class D fire extinguisher should be used for sodium fires.

Step-by-step solution

1) Reaction 1: 4 Na(s) + O2(g) → 2 Na2O(s)

First, find the standard enthalpies of formation (∆Hf°) for each substance in Appendix 4: Na(s): 0 kJ/mol (element in standard state) O2(g): 0 kJ/mol (element in standard state) Na2O(s): -416 kJ/mol Next, apply the formula to calculate ∆H° for the reaction: ∆H° = [2(∆Hf° of Na2O)] - [4(∆Hf° of Na) + 1(∆Hf° of O2)] ∆H° = [2(-416)] - [4(0) + 0] ∆H° = -832 kJ/mol

2) Reaction 2: 2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)

Find ∆Hf° for each substance in Appendix 4: Na(s): 0 kJ/mol (element in standard state) H2O(l): -285.8 kJ/mol NaOH(aq): -468.1 kJ/mol H2(g): 0 kJ/mol (element in standard state) Calculate ∆H° for the reaction: ∆H° = [2(∆Hf° of NaOH) + 1(∆Hf° of H2)] - [2(∆Hf° of Na) + 2(∆Hf° of H2O)] ∆H° = [2(-468.1) + 0] - [2(0) + 2(-285.8)] ∆H° = (-936.2) - (-571.6) ∆H° = -364.6 kJ/mol

3) Reaction 3: 2 Na(s) + CO2(g) → Na2O(s) + CO(g)

Find ∆Hf° for each substance in Appendix 4: Na(s): 0 kJ/mol (element in standard state) CO2(g): -393.5 kJ/mol Na2O(s): -416 kJ/mol CO(g): -110.5 kJ/mol Calculate ∆H° for the reaction: ∆H° = [1(∆Hf° of Na2O) + 1(∆Hf° of CO)] - [2(∆Hf° of Na) + 1(∆Hf° of CO2)] ∆H° = [(-416) + (-110.5)] - [2(0) + (-393.5)] ∆H° = (-526.5) - (-393.5) ∆H° = -133 kJ/mol In the first reaction, sodium reacts with oxygen, generating a large amount of heat. This suggests that sodium fires burn extremely hot, which may make it difficult to extinguish. Water is not suitable to extinguish sodium fires because of the second reaction. Water reacts with sodium to produce sodium hydroxide and hydrogen gas. Hydrogen gas is combustible, so using water to extinguish a sodium fire may make it more intense or even explosive. Using carbon dioxide as the fire extinguisher may not be effective due to the third reaction. Sodium can react with carbon dioxide to form sodium oxide and carbon monoxide, both of which do not help in putting out the fire. Considering these results, neither water nor carbon dioxide fire extinguishers are suitable for extinguishing sodium fires. A special Class D fire extinguisher should be used to effectively put out a sodium fire.

Key concepts

Enthalpy of Formation

The enthalpy of formation, denoted as \( \Delta H^\circ_f \), is a critical concept in thermochemistry. It's defined as the heat change that occurs when one mole of a compound is formed from its elements in their standard states. For instance, when sodium (\