STEM Problem Detail

Problem Statement

Prompt

provide me a step by step solution for problem 41.

Solution Rendering...

**Problem:**
Calculate the function
$$ \dfrac{G^{\circ} - H_0^{\circ}}{T} $$
for hydrogen chloride (HCl) at $$T = 298\ \text{K}$$ and
$$P = 1.0133 \times 10^{5}\ \text{Pa}$$.

Given data:
Internuclear distance
$$ r_e = 1.2715 \times 10^{-10}\ \text{m} $$

Vibrational frequency
$$ \tilde{\nu} = 2.9897 \times 10^{5}\ \text{m}^{-1} $$

Electronic partition function
$$ q_{\text{elec}} = 1 $$

Also determine $$(G^{\circ}_{298} - H^{\circ}_{298})$$.

**Theory:**
From statistical thermodynamics, the Gibbs free energy is related to
the molecular partition function $$q$$ by

$$ G^{\circ} - H_0^{\circ} = -RT \ln q $$

Hence,

$$ \dfrac{G^{\circ} - H_0^{\circ}}{T} = -R \ln q $$

For a diatomic ideal gas,

$$ q = q_{\text{trans}}\, q_{\text{rot}}\, q_{\text{vib}}\, q_{\text{elec}} $$

**Step 1: Translational partition function**

$$ q_{\text{trans}} =
\left(\dfrac{2\pi m k T}{h^2}\right)^{3/2}
\dfrac{kT}{P} $$

Molar mass of HCl:

$$ M = 36.46\ \text{g mol}^{-1} $$

$$ m = \dfrac{36.46 \times 10^{-3}}{6.022 \times 10^{23}}
= 6.05 \times 10^{-26}\ \text{kg} $$

Substituting values,

$$ q_{\text{trans}} \approx 2.6 \times 10^{6} $$

**Step 2: Rotational partition function**

For a heteronuclear diatomic molecule,

$$ q_{\text{rot}} = \dfrac{T}{\theta_{\text{rot}}} $$

The rotational temperature is

$$ \theta_{\text{rot}} =
\dfrac{h^2}{8\pi^2 I k} $$

Moment of inertia:

$$ I = \mu r_e^2 $$

Reduced mass:

$$ \mu =
\dfrac{m_{\text{H}} m_{\text{Cl}}}
{m_{\text{H}} + m_{\text{Cl}}}
\approx 1.63 \times 10^{-27}\ \text{kg} $$

$$ I = (1.63 \times 10^{-27})(1.2715 \times 10^{-10})^2
= 2.64 \times 10^{-47}\ \text{kg m}^2 $$

$$ \theta_{\text{rot}} \approx 15.2\ \text{K} $$

Thus,

$$ q_{\text{rot}} = \dfrac{298}{15.2} \approx 19.6 $$

**Step 3: Vibrational partition function**

Vibrational temperature:

$$ \theta_{\text{vib}} =
\dfrac{hc\tilde{\nu}}{k} $$

$$ \theta_{\text{vib}} =
\dfrac{(6.626\times10^{-34})(3\times10^{8})(2.9897\times10^{5})}
{1.381\times10^{-23}}
\approx 4300\ \text{K} $$

$$ q_{\text{vib}} =
\dfrac{1}{1 - \exp\!\left(-\dfrac{\theta_{\text{vib}}}{T}\right)} $$

Since $$\theta_{\text{vib}} \gg T$$,

$$ q_{\text{vib}} \approx 1 $$

**Step 4: Electronic partition function**

$$ q_{\text{elec}} = 1 $$

**Step 5: Total partition function**

$$ q = (2.6 \times 10^{6})(19.6)(1)(1) $$

$$ q \approx 5.1 \times 10^{7} $$

$$\textbf{Step 6: Evaluate } \dfrac{G^{\circ} - H_0^{\circ}}{T}$$

$$ \dfrac{G^{\circ} - H_0^{\circ}}{T}
= -R \ln q $$

$$ = -8.314 \ln(5.1 \times 10^{7}) $$

$$ = -8.314 \times 17.75 $$

**Final Answer (Part 1):**

$$ \boxed{\,\dfrac{G^{\circ} - H_0^{\circ}}{T}
\approx -148\ \text{J mol}^{-1}\text{K}^{-1}\,} $$

$$\textbf{Step 7: Evaluate } (G^{\circ}_{298} - H^{\circ}_{298})$$

$$ G^{\circ}_{298} - H^{\circ}_{298}
= T \left(\dfrac{G^{\circ} - H_0^{\circ}}{T}\right) $$

$$ = 298 \times (-148) $$

**Final Answer (Part 2):**

$$ \boxed{\,G^{\circ}_{298} - H^{\circ}_{298}
\approx -4.41 \times 10^{4}\ \text{J mol}^{-1}\,} $$

Loading solution...

Problem #51