Problem 147 - General Chemistry

Question

How much heat must be added to \(12.5 \mathrm{~g}\) of solid white phosphorus, \(\mathrm{P}_{4}\), at \(25.0^{\circ} \mathrm{C}\) to give the liquid at its melting point, \(44.1^{\circ} \mathrm{C} ?\) The heat capacity of solid white phosphorus is \(95.4 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})\); its heat of fusion is \(2.63 \mathrm{~kJ} / \mathrm{mol}\)

Step-by-Step Solution

Answer

449.67 J of heat is required to melt 12.5 g of phosphorus.

Step 1: Convert Mass to Moles

First, we need to convert the given mass of phosphorus into moles. The molar mass of white phosphorus (P_{4}) is approximately 123.88 g/mol. We use the formula: moles = \frac{mass}{molar \space mass}. Given mass is 12.5 g, so dividing by 123.88, we get \( n = \frac{12.5}{123.88} \approx 0.101 \text{ mol} \)

Step 2: Calculate Heat Required to Raise Temperature to Melting Point

Using the formula \( q = mc\Delta T \), where \(m\) is the number of molesconverted in step 1,\(c\) is the heat capacity, \(\Delta T = T_{m} - T_{i}\) is the change in temperature. Plugging in the values: \( c = 95.4 \, J/(K \cdot mol) \), \( \Delta T = 44.1 \degree C - 25.0 \degree C = 19.1 K \): \( q = 0.101 \, mol \times 95.4 \, J/(K \cdot mol) \times 19.1 \, K = 184.0434 \, J \).

Step 3: Calculate Heat Required for Fusion

Calculate the heat required to melt the phosphorus at its melting point using the formula \( q = n \cdot \Delta H_{fus} \). \( \Delta H_{fus} = 2.63 \, kJ/mol \) = 2630 J/mol, and \( n = 0.101 \, mol \) from step 1. Substituting these values, \( q = 0.101 \, mol \times 2630 \, J/mol = 265.63 \, J \).

Step 4: Total Heat Required

Add the heat calculated in Step 2 and Step 3 to determine the total heat required. Total heat \[ q_{total} = 184.0434 \, J + 265.63 \, J = 449.6734 \, J \].

Key Concepts

Heat Capacity

Heat capacity is an essential concept in understanding how substances absorb and release energy. It is defined as the amount of heat required to change the temperature of a substance by one degree Celsius (or Kelvin). In the context of this exercise, we focused on the **specific heat capacity** of phosphorus, which is denoted as 95.4 J/(K·mol). This value tells us how much energy is needed to raise the temperature of one mole of phosphorus by one degree.

• **Molar Heat Capacity**: In this case, we're considering the molar heat capacity, because the calculation is per mole of phosphorus, rather than per gram. This is why converting the mass of phosphorus to moles is a crucial step.
• **Energy Transfer**: Understanding heat capacity allows us to predict how much energy is transferred during heating or cooling processes.

Comprehending heat capacity is crucial when calculating the energy needed to change the temperature of a given amount of substance before it undergoes a phase change, such as melting.

Heat of Fusion

The heat of fusion is the amount of energy required to change a solid into a liquid at its melting point, without changing its temperature. This energy goes into overcoming the forces that hold the solid structure together, enabling it to transition into the liquid state.

• **Latent Heat**: This concept is known as latent heat because the energy change does not reflect as a temperature increase.
• **Heat of Fusion for Phosphorus**: In our exercise, the given heat of fusion for phosphorus is 2.63 kJ/mol. This means each mole of phosphorus needs this amount of heat to transition from solid to liquid.
• **Phase Equilibrium**: At the melting point, the substance is in a state of phase equilibrium, where both solid and liquid phases coexist.

Knowing the heat of fusion is critical to understanding how much heat is involved purely in the phase change process.

Phase Change

A phase change refers to the transformation of a substance from one state of matter to another, such as solid to liquid, liquid to gas, or vice versa. In our context, we are looking at the transition of solid phosphorus into its liquid form, known as melting.

• **Temperature Invariance**: During a phase change, the temperature of the substance remains constant. The heat added is used solely to alter the phase, not to change temperature.
• **Energy Absorption and Release**: During melting, energy is absorbed to break the bonds in the rigid structure of the solid, allowing the particles to move more freely as a liquid.
• **Melting Point of Phosphorus**: For phosphorus, this occurs at 44.1°C, which is its specific melting point where the phase change is taking place.

Understanding phase changes helps us predict and calculate the energy requirements for materials undergoing state transformations under various conditions.

Temperature Conversion

Temperature conversion is a fundamental skill in chemistry and physics, allowing us to compare and analyze data across different temperature scales. In scientific calculations, Kelvin is often used alongside Celsius, as they are directly convertible.

• **Kelvin and Celsius**: The conversion is straightforward; Celsius to Kelvin involves adding 273.15 to the Celsius temperature, making it simple to convert between these scales.
• **Why Convert?**: In thermodynamics, Kelvin is preferred because it is an absolute temperature scale, beginning at absolute zero, where molecular motion ceases.
• **Example Calculation in Exercise**: While our exercise used Celsius directly for calculating temperature change (ΔT), knowing the conversion process is crucial, especially when dealing with different heat capacity units or broader scientific contexts.

Mastering temperature conversion procedures allows accurate energy calculations across various contexts and ensures clarity in communication and comprehension of thermodynamic processes.