Question

The value of enthalpy change, H_β − H_α (in J mol⁻¹), at 300 K is ___.

Answer 1

Correct Answer: 300

The enthalpy change equation is given as:

\(\Delta H_{600} - \Delta H_{300} = 1 \times (C_{p,\beta} - C_{p,\alpha}) (600 - 300)\)

Now, at the transition temperature, we have:

\(\Delta H_{600} = T \Delta S_{600}\)

Substitute the given values:

\(600 = 600 \times (6 - 5) = 600 \, \text{J mol}^{-1}\)

From the equation:

\(600 - \Delta H_{300} = 1 \times 1 \times 300\)

Solving for \( \Delta H_{300} \):

\(\Delta H_{300} = 600 - 300 = 300 \, \text{J mol}^{-1}\)

Thus, the enthalpy change at 300 K is: \( \Delta H_{300} = 300 \, \text{J mol}^{-1} \).

Answer 2

The equation for the entropy change is given as:

\[\Delta S_{600} - \Delta S_{300} = \int_{300}^{600} \frac{1 \times (C_{p,\beta} - C_{p,\alpha})}{T} dT\]

Next, this simplifies to:

\(= 1 \times \int_{300}^{600} \ln \left( \frac{T_2}{T_1} \right) dT \)

Where:

• T2 = 600 K
• T1 = 300 K

Now, performing the integration:

\(\Delta S_{300} = 1 \times \ln \left( \frac{600}{300} \right)\)

This simplifies to:

\(1 - \Delta S_{300} = 1 \times \ln 2\)

Thus:

\(\Delta S_{300} = 1 - 0.69 = 0.31 \, \text{J mol}^{-1} \text{K}^{-1}\)

Therefore, the entropy change at 300 K is: \(\Delta S_{300} = 0.31 \, \text{J mol}^{-1} \text{K}^{-1}\).