Question

The heat capacity is critical in calculating the enthalpy and entropy. Using the data (heat capacity) from book or internet and also ∆fus H (I2) = 15.52387kJ/mol at 387K and ∆vap (I2) =41.57kJ/mol at 457K. a) Calculate the heat required to bring one mole of iodine from 0K to 500k. b) calculate the absolute entropy of iodine at 500K

Answer 1

Using the data (heat capacity) from book or internet and also $\Delta H_{fier}(\mathrm{I}_2) = 15.524\mathrm{kJ / mol}$ at $387~\mathrm{K}$ and $\Delta H_{vap}(\mathrm{I}_2) = 41.57\mathrm{kJ / mol}$ at $457~\mathrm{K}$ : a) calculate the heat required to bring one mole of iodine from $0\mathrm{K}$ to $500\mathrm{K}$ ; b) calculate the absolute entropy of iodine at $500\mathrm{K}$ .

Solution: Temperature dependencies of molar heat capacity $(\mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1})$ of iodine are:

1) in solid state, $C_{p(x)}^{\circ}(T) = 40.12 + 0.0498 \cdot T; T = 298 - 387 \mathrm{~K}$ ;

According to the Debye $T^{\delta}$ law, dependency of the heat capacity of solids at the low temperatures can be shown as $C_{p(x)}^{\circ}(T) = k \cdot T^{3}$ ; $T = 0 - 15\mathrm{K}$ .

The coefficient $k$ can be calculated as: $k = \frac{C_p^\circ(15)}{15^3} = \frac{9.2}{3375} = 2.726 \cdot 10^{-3} \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}^4}$ .

2) in gaseous state, $C_{p(g)}^{\circ}(T) = 37.4 + 5.9 \cdot 10^{-4} \cdot T - \frac{7.1 \cdot 10^{4}}{T^{2}}$ ; $T = 298 - 1000 \mathrm{~K}$ ;

As you know, iodine sublimes (becomes a gas directly from the solid phase) instead of melting. Latent heat of sublimation is equal to the sum of latent heats of fusion and vaporization:

$$\Delta H _ {s u b} \left(I _ {2}\right) = 1 5. 5 2 4 + 4 1. 5 7 = 5 7. 0 9 4 \mathrm {k J / m o l} = 5 7, 0 9 4 \mathrm {J / m o l}.$$

a) The amount of heat $Q$ , required for the heating of one mole of iodine from 0 K to 500 K at constant pressure (1 atm) can be calculated as the change of standard molar enthalpy, which depends from the heat capacity of iodine and heat of phase transition (latent heat of sublimation):

$Q = \Delta H_{500}^{\circ} = \int_{0}^{T_{s}} C_{p(s)}^{\circ}(T) dT + \Delta H_{sub} + \int_{T_{s}}^{500} C_{p(g)}^{\circ}(T) dT$ , where $T_{s}$ is the temperature of sublimation of iodine, which is equal to the melting point (387 K).

As you see, dependency of molar heat capacity of solid iodine in the temperature range from $15\mathrm{K}$ to $300\mathrm{K}$ is available only in the tabular form. Thus, we must calculate the value of $\int_{15}^{300}C_{p(s)}^{\circ}(T)dT$ by graphical integration, which is based on determining of the area under the curve $C_{p(s)}^{\circ}(T)$ from $15\mathrm{K}$ to $300\mathrm{K}$ .

Then, $Q = \int_{0}^{15} C_{p(s)}^{\circ}(T) dT + (H_{300}^{\circ} - H_{15}^{\circ}) + \int_{300}^{387} C_{p(s)}^{\circ}(T) dT + \Delta H_{sub} + \int_{387}^{500} C_{p(g)}^{\circ}(T) dT$ , where $(H_{300}^{\circ} - H_{15}^{\circ})$ is the value, determined by graphical integration.

First integral is equal to: $\int_{0}^{15} C_{p(s)}^{\circ}(T) dT = \int_{0}^{15} k \cdot T^{3} dT = \frac{k}{4} \cdot T^{4}\bigg|_{0}^{15} = \frac{2.726 \cdot 10^{-3}}{4} \cdot 15^{4} = 34.5 \frac{\mathrm{J}}{\mathrm{mol}}$ .

Graphical integration gives the value of $\int_{15}^{300} C_{p(s)}^{\circ}(T) dT = (H_{300}^{\circ} - H_{15}^{\circ}) = 13315.25 \frac{\mathrm{J}}{\mathrm{mol}}$ (graph of the function $C_{p(s)}^{\circ} = f(T)$ is shown below).

And the rest of integrals are: $\int_{300}^{387} C_{p(x)}^*(T) dT = \int_{300}^{387} (40.12 + 0.0498 \cdot T) dT = \left(40.12 \cdot T + \frac{0.0498}{2} \cdot T^2\right)\bigg|_{300}^{387} =$

$= 40.12\cdot (387 - 300) + 0.0249\cdot (387^{2} - 300^{2}) = 4978.7\frac{\mathrm{J}}{\mathrm{mol}}$

$\int_{387}^{500}C_{p(g)}^{*}(T)dT = \int_{387}^{500}\left(37.4 + 5.9\cdot 10^{-4}\cdot T - \frac{7.1\cdot 10^{4}}{T^{2}}\right)dT = \left(37.4\cdot T + \frac{5.9\cdot 10^{-4}}{2}\cdot T^{2} + \frac{7.1\cdot 10^{4}}{T}\right)\bigg|_{387}^{500} =$

$= 37.4\cdot (500 - 387) + 2.95\cdot 10^{-4}\cdot (500^{2} - 387^{2}) + 7.1\cdot 10^{4}\cdot \left(\frac{1}{500} -\frac{1}{387}\right) = 4214.3\frac{\mathrm{J}}{\mathrm{mol}}.$

Then, $Q = 34.5 + 13315.25 + 4978.7 + 57,094 + 4214.3 = 79636.75$ J = 79.64 kJ.

Answer: 79.64 kJ.

b) The absolute entropy of iodine at $500\mathrm{K}$ can be calculated as the standard entropy of solid iodine at $298\mathrm{K}$ plus change of absolute entropy in the range from $298\mathrm{K}$ to $500\mathrm{K}$ , which depends from the heat capacity of iodine and heat of phase transition (latent heat of sublimation):

$S_{500}^{\circ} = S_{298}^{\circ} + \Delta S_{500}^{\circ} = S_{298}^{\circ} + \int_{298}^{T_S}\frac{C_{p(x)}^{\circ}(T)}{T} dT + \frac{\Delta H_{sub}}{T_S} +\int_{T_S}^{500}\frac{C_{p(g)}^{\circ}(T)}{T} dT,$ where $S_{298}^{\circ} = 116.14\frac{\mathrm{J}}{\mathrm{mol}\cdot\mathrm{K}}$ is the standard entropy of solid iodine at $298\mathrm{K}$ .

First integral is equal to: $\int_{298}^{T_S}\frac{C_{p(x)}^*(T)}{T} dT = \int_{298}^{387}\frac{40.12 + 0.0498\cdot T}{T} dT = \int_{298}^{387}\left(\frac{40.12}{T} +0.0498\right)dT =$

$= (40.12\cdot \ln T + 0.0498\cdot T)\bigg|_{298}^{387} = 40.12\cdot \ln \frac{387}{298} +0.0498\cdot (387 - 298) = 14.92\frac{\mathrm{J}}{\mathrm{mol}\cdot\mathrm{K}}.$ And the second integral is: $\int_{T_S}^{500}\frac{C_{p(g)}^*(T)}{T} dT = \int_{387}^{500}\left(\frac{37.4}{T} +5.9\cdot 10^{-4} - \frac{7.1\cdot 10^4}{T^3}\right)dT = \left(37.4\cdot \ln T + 5.9\cdot 10^{-4}\cdot T + \frac{7.1\cdot 10^4}{2\cdot T^2}\right)\bigg|_{387}^{500} =$

$= 37.4\cdot \ln \frac{500}{387} +5.9\cdot 10^{-4}\cdot (500 - 387) + 3.55\cdot 10^{4}\cdot \left(\frac{1}{500^{2}} -\frac{1}{387^{2}}\right) = 9.55\frac{\mathrm{J}}{\mathrm{mol}\cdot\mathrm{K}}.$

Then, absolute entropy of iodine at $500\mathrm{K}$ is $S_{500}^{\circ} = 116.14 + 14.92 + \frac{57,094}{387} + 9.55 = 288.14\frac{\mathrm{J}}{\mathrm{mol}\cdot\mathrm{K}}$ .

Answer: 288.14 J/(mol·K).