Question

Computing enthalpy and entropy with a temperaturedependent heat capacity. The heat capacity for liquid \(n\)-butane depends on temperature: $$ C_{p}(T)=a+b T, $$ where \(a=100 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) and \(b=0.1067 \mathrm{JK}^{-2} \mathrm{~mol}^{-1}\), from its freezing temperature \(T_{f} \approx 140 \mathrm{~K}\) to \(T_{b} \approx 270 \mathrm{~K}\), its boiling temperature. (a) Compute \(\Delta H\) for heating liquid butane from \(T_{A}=\) \(170 \mathrm{~K}\) to \(T_{B}=270 \mathrm{~K}\). (b) Compute \(\Delta S\) for the same process.

Short answer

ΔH = 12345.735 Jmol^{-1}; ΔS = 56.932 Jmol^{-1}K^{-1}

Step-by-step solution

Identify Given Data

Identify the given data for the problem: Heat capacity formula: \[ C_{p}(T) = a + bT \]with constants,\[ a = 100 \text{ JK}^{-1}\text{mol}^{-1} \]\[ b = 0.1067 \text{ JK}^{-2}\text{mol}^{-1} \]Temperature range from \(T_A = 170 \text{ K}\) to \(T_B = 270 \text{ K}\).

Compute Enthalpy Change (ΔH)

To compute the enthalpy change \( \Delta H \), integrate the heat capacity with respect to temperature: \[ \Delta H = \int_{T_{A}}^{T_{B}} C_p(T) \, dT \]Substituting \( C_p(T) = a + bT \):\[ \Delta H = \int_{170}^{270} (100 + 0.1067T) \, dT \]Integrate each term separately:\[ \Delta H = \int_{170}^{270} 100 \, dT + \int_{170}^{270} 0.1067T \, dT \]\[ \Delta H = 100[T]_{170}^{270} + 0.1067[\frac{T^2}{2}]_{170}^{270} \]

Evaluate Integrals for ΔH

Evaluate the first integral: \[ 100[T]_{170}^{270} = 100(270 - 170) = 100(100) = 10000 \text{ Jmol}^{-1} \]Evaluate the second integral:\[ 0.1067[\frac{T^2}{2}]_{170}^{270} = 0.1067[\frac{270^2}{2} - \frac{170^2}{2}] \]Solve for each term:\[ 0.1067 \frac{270^2}{2} = 0.1067 \frac{72900}{2} = 0.1067(36450) = 3888.15 \text{ Jmol}^{-1} \]\[ 0.1067 \frac{170^2}{2} = 0.1067 \frac{28900}{2} = 0.1067(14450) = 1542.415 \text{ Jmol}^{-1} \]Subtraction gives:\[ 3888.15 - 1542.415 = 2345.735 \text{ Jmol}^{-1} \]

Find Total ΔH

Combine the results from Step 3:\[ \Delta H = 10000 + 2345.735 = 12345.735 \text{ Jmol}^{-1} \]Therefore,\[ \Delta H = 12345.735 \text{ Jmol}^{-1} \]

Compute Entropy Change (ΔS)

To find the entropy change \( \Delta S \), use the formula:\[ \Delta S = \int_{T_{A}}^{T_{B}} \frac{C_p(T)}{T} \, dT \]Substitute \( C_p(T) = a + bT \):\[ \Delta S = \int_{170}^{270} \frac{100 + 0.1067T}{T} \, dT \]Split the integral into two parts:\[ \Delta S = \int_{170}^{270} \frac{100}{T} \, dT + \int_{170}^{270} 0.1067 \, dT \]Integrate each term separately:\[ \Delta S = 100[\ln(T)]_{170}^{270} + 0.1067[T]_{170}^{270} \]

Evaluate Integrals for ΔS

Evaluate the first integral:\[ 100[\ln(T)]_{170}^{270} = 100(\ln(270) - \ln(170)) \]\[ 100( 5.59842 - 5.1358) = 100(0.46262) = 46.262 \text{ Jmol}^{-1}K^{-1} \]Evaluate the second integral:\[ 0.1067[T]_{170}^{270} = 0.1067(270 - 170) = 0.1067(100) = 10.67 \text{ Jmol}^{-1}K^{-1} \]

Find Total ΔS

Combine the results from Step 6:\[ \Delta S = 46.262 + 10.67 = 56.932 \text{ Jmol}^{-1}K^{-1} \]Therefore,\[ \Delta S = 56.932 \text{ Jmol}^{-1}K^{-1} \]

Key concepts

Enthalpy Change

Enthalpy change (\(\Delta H\)) is the heat absorbed or released during a process at constant pressure. For a temperature-dependent heat capacity like in the case of liquid butane, we need to integrate the heat capacity (\(C_p(T)\)) over the temperature range of interest.
It’s given that \(C_p(T) = a + bT\) with \(a = 100 \text{ JK}^{-1}\text{mol}^{-1}\) and \(b = 0.1067 \text{ JK}^{-2}\text{mol}^{-1}\).

The enthalpy change is then calculated using the integral: \[ \Delta H = \int_{T_{A}}^{T_{B}} C_p(T) \, dT \] For this problem, we integrate: \[ \Delta H = \int_{170}^{270} (100 + 0.1067T) \, dT \] By solving the integral step by step, we get: \[ \Delta H = 10000 + 2345.735 = 12345.735 \text{ Jmol}^{-1} \]
So, the change in enthalpy for heating from 170 K to 270 K is \(\Delta H = 12345.735 \text{ Jmol}^{-1}\).

Entropy Change

Entropy change (\(\Delta S\)) measures the disorder or randomness in a system. In our problem, we use the formula for entropy change when the heat capacity depends on temperature: \[ \Delta S = \int_{T_{A}}^{T_{B}} \frac{C_p(T)}{T} \, dT \]
Since \(C_p(T) = 100 + 0.1067T \, \, \, \, \), we split the integral into two parts: \[ \Delta S = \int_{170}^{270} \frac{100}{T} \, dT + \int_{170}^{270} 0.1067 \, dT \]
The first part deals with the natural logarithm, while the second part involves a regular integral of a constant.

Evaluating these integrals, we find: \[ 100[\ln(T)]_{170}^{270} = 46.262 \text{ Jmol}^{-1}K^{-1} \] \[ 0.1067[T]_{170}^{270} = 10.67 \text{ Jmol}^{-1}K^{-1} \]
Adding them gives the total entropy change: \[ \Delta S = 46.262 + 10.67 = 56.932 \text{ Jmol}^{-1}K^{-1} \]
Therefore, the change in entropy for the process is \(\Delta S = 56.932 \text{ Jmol}^{-1}K^{-1}\).

Temperature-Dependent Heat Capacity

Heat capacity (\(C_p(T)\)) is the amount of heat required to change the temperature of a substance by one degree. For substances like liquid butane, the heat capacity changes with temperature and can be expressed as a linear equation: \[ C_p(T) = a + bT \] where \(a\) and \(b\) are constants. In our example, \(a = 100 \text{ JK}^{-1}\text{mol}^{-1}\) and \(b = 0.1067 \text{ JK}^{-2}\text{mol}^{-1}\).

Having a temperature-dependent heat capacity means that energy added to or removed from the substance affects its temperature in a nonlinear way. This requires integration to determine the changes in enthalpy or entropy over temperature ranges.

Such calculations are crucial in thermodynamics as they help predict how a substance will behave when subjected to temperature changes in real-world scenarios. This information is vital in chemical engineering and physical chemistry applications, ensuring processes are efficient and safe.