Problem 83 - Chemistry: An Atoms-Focused Approach

Question

Show that hydrogen cyanide (HCN) is a gas at \(25^{\circ} \mathrm{C}\) by estimating its normal boiling point from the following data:$$\begin{array}{ccc} & \Delta H_{i}^{\circ}(\mathrm{k} J / \mathrm{mol}) & S^{\circ}[J /(\mathrm{mol} \cdot \mathrm{K})] \\\\\mathrm{HCN}(\ell) & 108.9 & 113 \\ \hline \mathrm{HCN}(g) & 135.1 & 202 \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

Based on the given enthalpy and entropy data, the normal boiling point of hydrogen cyanide (HCN) is calculated to be 294 K. Since this boiling point is lower than the temperature of 25°C (298.15 K), we can conclude that HCN is a gas at 25°C.

Step 1: Calculate the Changes in Enthalpy and Entropy

First, we need to determine the changes in enthalpy \((\Delta H^{\circ})\) and entropy \((\Delta S^{\circ})\) for the phase transition from liquid to gas. $$\Delta H^{\circ} = \Delta H^{\circ}_{g} - \Delta H^{\circ}_{l} = 135.1\;\mathrm{kJ/mol} - 108.9\;\mathrm{kJ/mol} = 26.2\;\mathrm{kJ/mol}$$ $$\Delta S^{\circ} = S^{\circ}_{g} - S^{\circ}_{l} = 202\;\mathrm{J/(mol\cdot K)} - 113\;\mathrm{J/(mol\cdot K)} = 89\;\mathrm{J/(mol\cdot K)}$$

Step 2: Calculate the Boiling Point

Since the \(\Delta G^{\circ} = 0\) at the boiling point, we can use the Gibbs free energy equation to find the boiling point temperature. $$0 = \Delta H^{\circ} - T\Delta S^{\circ}$$ Rearranging for \(T\), we get: $$T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}$$ Plugging in the values we calculated in step 1, we have: $$T = \frac{26.2\;\mathrm{kJ/mol}}{89\;\mathrm{J/(mol\cdot K)}} = \frac{26.2 \times 10^3\;\mathrm{J/mol}}{89\;\mathrm{J/(mol\cdot K)}} = 294\;\mathrm{K}$$

Step 3: Comparing Boiling Point and Given Temperature

Now we need to compare the calculated boiling point with the given temperature of \(25^{\circ} \mathrm{C}\). First, convert \(25^{\circ} \mathrm{C}\) to Kelvin: $$T_{25} = 25^{\circ} \mathrm{C} + 273.15 = 298.15\;\mathrm{K}$$ As we can see, the boiling point of HCN (294 K) is lower than \(25^{\circ} \mathrm{C}\) (298.15 K). Therefore, it can be concluded that hydrogen cyanide (HCN) is a gas at \(25^{\circ} \mathrm{C}\).

Key Concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is an essential concept in thermodynamics. It represents the amount of energy absorbed or released as heat during a chemical reaction at constant pressure. Here, we are exploring the phase transition of hydrogen cyanide from liquid (HCN(ℓ)) to gas (HCN(g)).

• The enthalpy change \( (\Delta H^{\circ}) \) for this transition is calculated by subtracting the enthalpy of the liquid from that of the gas: \[ \Delta H^{\circ} = \Delta H^{\circ}_{g} - \Delta H^{\circ}_{l} = 135.1\;\mathrm{kJ/mol} - 108.9\;\mathrm{kJ/mol} = 26.2\;\mathrm{kJ/mol} \]
• This positive \( \Delta H^{\circ} \) indicates that energy is required for the phase transition from liquid to gas, which is endothermic.
• The magnitude signifies how much energy is involved in overcoming intermolecular forces in the liquid to achieve the gaseous state.

Understanding the enthalpy change helps in predicting whether substances will absorb or release heat, a vital factor in determining the feasibility and energy requirements of chemical processes.

Entropy Change

Entropy change, symbolized as \( \Delta S \), measures the degree of disorder or randomness in a system. It sheds light on the distribution of energy and the dispersal of matter in a chemical reaction or physical transformation.

• For hydrogen cyanide's phase change from liquid to gas, the entropy change is calculated as follows: \[ \Delta S^{\circ} = S^{\circ}_{g} - S^{\circ}_{l} = 202\;\mathrm{J/(mol \cdot K)} - 113\;\mathrm{J/(mol \cdot K)} = 89\;\mathrm{J/(mol \cdot K)} \]
• A positive \( \Delta S^{\circ} \) signifies an increase in randomness or disorder as the liquid transitions to gas, which is expected because gas molecules exhibit more freedom and disorder compared to liquids.
• This concept is key in understanding spontaneity as systems tend to move towards higher entropy.

Entropy change helps us comprehend why certain processes occur naturally and how energy distributes across chemical systems, often favoring states of higher disorder.

Gibbs Free Energy

Gibbs free energy, expressed as \( \Delta G \), is a pivotal concept in predicting the spontaneity of reactions and phase changes. It integrates enthalpy and entropy changes to provide insights about system feasibility under constant pressure and temperature.

• The equation \( \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \) helps determine the point at which processes become spontaneous, specifically the temperature for phase transitions like boiling.
• For the boiling point calculation, when \( \Delta G^{\circ} = 0 \), it suggests equilibrium between the liquid and gas phases. Solving for temperature \( T \) gives the boiling point: \[ T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}} = \frac{26.2 \times 10^3\;\mathrm{J/mol}}{89\;\mathrm{J/(mol \cdot K)}} = 294\;\mathrm{K} \]
• At this temperature, the transition is feasible, explaining why HCN is a gas at room temperature, as \( 294\;\mathrm{K} \) is less than the external temperature of \( 298.15\;\mathrm{K} \).

Gibbs free energy is a comprehensive tool for understanding not just the conditions for spontaneity but also exploring energy efficiency and stability in different states.