Question

The heat capacity of solid lead oxide is given by \\[ C_{P, m}=44.35+1.47 \times 10^{-3} \frac{T}{\mathrm{K}} \text { in units of } \mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1} \\] Calculate the change in enthalpy of 1.75 mol of \(\mathrm{PbO}(s)\) if it is cooled from 825 K to 375 K at constant pressure.

Short answer

The change in enthalpy for 1.75 mol of solid lead oxide (PbO) cooled from 825 K to 375 K at constant pressure is approximately -78,251.5 J.

Step-by-step solution

Write down the formula for the change in enthalpy.

The change in enthalpy ΔH is given by the integral of the constant-pressure molar heat capacity (Cp,m) with respect to temperature (T): \[ \Delta H = \int_{T_1}^{T_2} C_{P,m}(T) dT \] where \(T_1\) is the initial temperature and \(T_2\) is the final temperature.

Rewrite the formula for heat capacity in terms of its variables

The heat capacity of solid lead oxide, \(C_{P,m}(T) \), is given as: \[ C_{P,m}(T)=44.35+1.47 \times 10^{-3} \frac{T}{\text{K}} \]

Perform integration

Now we substitute the given heat capacity formula into the enthalpy integral: \[ \Delta H = \int_{T_1}^{T_2} \left(44.35 + 1.47 \times 10^{-3} \frac{T}{\text{K}}\right) dT \] Integrating with respect to temperature, we get: \[ \Delta H = \left[44.35T + \frac{1}{2} \cdot 1.47 \times 10^{-3} \frac{T^2}{\text{K}}\right]_{T_1}^{T_2} \]

Substitute the temperature values

Now, we substitute the initial temperature \(T_1 = 825\,\text{K}\) and final temperature \(T_2 = 375\, \text{K}\) in the formula: \[ \Delta H = \left[44.35(375\,\text{K}) + \frac{1}{2} \cdot 1.47 \times 10^{-3} \frac{(375\,\text{K})^2}{\text{K}}\right] - \left[44.35(825\,\text{K}) + \frac{1}{2} \cdot 1.47 \times 10^{-3} \frac{(825\,\text{K})^2}{\text{K}}\right] \]

Calculate the change in enthalpy

Performing the calculations, we obtain: \[ \Delta H = (-44743.7\, \text{J/mol}) \]

Multiply by the number of moles

Finally, multiply the change in enthalpy by the number of moles (1.75 mol) to get the overall change in enthalpy for the entire amount of solid lead oxide: \[ \Delta H_{\text{total}} = 1.75\, \text{mol} \times (-44743.7\, \text{J/mol}) = (-78251.5\, \text{J}) \] The change in enthalpy for 1.75 mol of solid lead oxide cooled from 825 K to 375 K at constant pressure is approximately -78,251.5 J.

Key concepts

Understanding Heat Capacity

Heat capacity is a crucial concept in thermodynamics, describing how much heat energy is needed to change the temperature of a substance. Simply put, it tells us how much energy is required to raise or lower the temperature of an object by 1 degree Kelvin. Heat capacity can be applied to both entire objects or specifically to one mole of substance, known as molar heat capacity.
The formula given in the exercise, \[ C_{P, m}=44.35+1.47 \times 10^{-3} \frac{T}{\mathrm{K}} \] describes how the heat capacity of lead oxide changes with temperature. This equation indicates that heat capacity not only depends on the material but also varies with temperature. As the temperature increases, the heat capacity slightly increases due to the linear term \(1.47 \times 10^{-3} T\) in the equation.
Understanding this varying nature of heat capacity is essential because it affects calculations in thermodynamics, especially when determining the change in enthalpy, as we do in the problem.

Integration in Thermodynamics

In thermodynamics, integration is a powerful tool used to calculate quantities like work, heat transfer, and changes in thermodynamic potentials, including enthalpy. When dealing with variable heat capacity, simply multiplying heat capacity by a temperature difference is not enough—you need to integrate the heat capacity function over the temperature range.
The integral formula from the exercise is:
\[ \Delta H = \int_{T_1}^{T_2} C_{P,m}(T) \, dT \]
This equation integrates the molar heat capacity \( C_{P,m} \) from an initial temperature \( T_1 \) to a final temperature \( T_2 \). It calculates the total change in enthalpy, considering the heat capacity throughout the temperature range, not just at a single point.
Performing this integration requires substitution of the heat capacity function, followed by the integration process, which in this case involves finding the antiderivative of each term in the equation. Integration allows for accurate calculations by taking into account how heat capacity varies with temperature.

Constant Pressure in Thermodynamics

When we talk about processes occurring at constant pressure in thermodynamics, it gives specific advantages and simplifies certain calculations. The assumption of constant pressure allows us to explore enthalpy changes directly without worrying about pressure variations affecting the results.
In the context of the given exercise, the calculation of enthalpy change assumes that the pressure remains constant while the lead oxide cools. This assumption is key because enthalpy, which represents the total heat content of a system, closely relates to work done by or on the system when a change occurs at constant pressure.
At constant pressure, the change in enthalpy accounts for not only the change in internal energy but also the energy needed for a system to perform work on its surroundings. This condition makes the enthalpy change calculable by integrating the heat capacity over the temperature range, as seen in the problem solution. Such assumptions are common in chemical processes and physical reactions where pressure remains stable.