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The molar heat of vaporization of sodium chloride is \(180 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} .\) Calculate the ratio of the vapor pressure of sodium chloride at \(1100^{\circ} \mathrm{C}\) to that at \(900^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The vapor pressure ratio at 1100°C to 900°C is approximately 21.3.

Step by step solution

01

Understand the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation relates the vapor pressure of a substance at two temperatures. It is expressed as: \[ \ln \left( \frac{P_2}{P_1} \right) = - \frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] where \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\), \(\Delta H_{vap}\) is the molar heat of vaporization, and \(R\) is the ideal gas constant, \(8.314 \, \text{J/mol K}\).
02

Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin for use in the Clausius-Clapeyron equation. The conversion formula is \(T(\text{K}) = T(\degree \text{C}) + 273.15\). Thus:- \(T_1 = 900 + 273.15 = 1173.15 \, \text{K}\)- \(T_2 = 1100 + 273.15 = 1373.15 \, \text{K}\)
03

Calculate the Ratio of Vapor Pressures

Substitute the given values into the Clausius-Clapeyron equation:\[ \ln \left( \frac{P_2}{P_1} \right) = - \frac{180,000}{8.314} \left( \frac{1}{1373.15} - \frac{1}{1173.15} \right) \]Calculate inside the parenthesis first:\[ \frac{1}{1373.15} - \frac{1}{1173.15} = -0.0001419 \, \text{K}^{-1} \]Then compute:\[ \ln \left( \frac{P_2}{P_1} \right) = - \frac{180,000}{8.314} \times -0.0001419\]\[ \ln \left( \frac{P_2}{P_1} \right) \approx 3.057 \]
04

Solve for Vapor Pressure Ratio

To find the ratio \( \frac{P_2}{P_1} \), exponentiate the result from the Clausius-Clapeyron equation:\[ \frac{P_2}{P_1} = e^{3.057} \approx 21.3 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vapor Pressure
Vapor pressure is a term that describes the pressure exerted by a vapor in equilibrium with its liquid or solid phase. Understanding vapor pressure is critical, especially when discussing processes like boiling or evaporation.
When a liquid vaporizes, its molecules escape from the surface into the air. If this occurs in a closed container, a dynamic equilibrium develops where molecules return to the liquid at the same rate they escape.

The pressure created at this dynamic equilibrium state is the vapor pressure, which depends on temperature and the strength of the intermolecular forces between molecules.
  • Stronger forces mean lower vapor pressure.
  • Higher temperature means higher vapor pressure due to increased molecular motion.
Vapor pressure is a key concept in physical chemistry and thermodynamics, explaining phenomena like the boiling point and even weather patterns. It's a critical aspect to consider in the Clausius-Clapeyron equation, which estimates how vapor pressure changes with temperature.
Molar Heat of Vaporization
The molar heat of vaporization, (\( \Delta H_{vap} \)), is the amount of heat energy required to vaporize one mole of a liquid substance at constant pressure.
It's a measure of the energy needed to overcome intermolecular forces and convert a liquid into a gas.

When dealing with the molar heat of vaporization, there are key points to consider:
  • A high (\( \Delta H_{vap} \)) usually indicates strong intermolecular forces within the liquid.
  • It is expressed in units of kilojoules per mole (kJ/mol).
  • It affects the rate of vaporization and the vapor pressure.
The value of (\( \Delta H_{vap} \)) also provides insights into a substance's boiling point and energy requirements for phase changes. Its importance is evident in various fields, such as chemistry and engineering, for designing processes like distillation.
Temperature Conversion
Temperature conversion is often necessary in scientific calculations, especially when using equations like the Clausius-Clapeyron equation that rely on absolute temperature scales.
The Kelvin scale is typically used in scientific contexts because it starts at absolute zero, ensuring non-negative temperature values.

Here's how to convert Celsius to Kelvin:
  • Add 273.15 to the Celsius temperature. For example, a temperature of 900°C is converted to 1173.15 K by calculating (\( 900 + 273.15 = 1173.15 \)).
  • The Kelvin scale is crucial for equations involving thermodynamic quantities.
Converting temperatures accurately ensures correct application of laws and equations in science. Whether dealing with various physicochemical calculations or understanding natural phenomena, using the Kelvin scale is essential for precise and meaningful results.

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Most popular questions from this chapter

Predict whether the entropy of the substance increases, decreases, or remains the same in the following processes: (a) \(\operatorname{Ar}(l) \rightarrow \operatorname{Ar}(g)\) (b) \(\mathrm{O}_{2}(g, 200 \mathrm{kPa}, 300 \mathrm{~K}) \rightarrow \mathrm{O}_{2}(g, 100 \mathrm{kPa}, 300 \mathrm{~K})\) (c) \(\mathrm{Cu}(s, 300 \mathrm{~K}) \rightarrow \mathrm{Cu}(s, 800 \mathrm{~K})\) (d) \(\mathrm{CO}_{2}(g) \rightarrow \mathrm{CO}_{2}(s)\)

For the equation \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \leftrightharpoons 2 \mathrm{NO}(g)\), use the following data to calculate the value of \(\Delta H_{\mathrm{rxan}}^{\circ}:\) \begin{tabular}{cc} \hline \(\boldsymbol{T} / \mathbf{K}\) & \(\boldsymbol{K}_{\mathrm{p}} / \mathbf{1 0}^{-4}\) \\ \hline 2000 & \(4.08\) \\ 2100 & \(6.86\) \\ 2200 & \(11.0\) \\ 2300 & \(16.9\) \\ 2400 & \(25.1\) \\ \hline \end{tabular} Compare your answer to the value calculated from the data in Appendix D.

At \(2000 \mathrm{~K}\) and one bar, water vapor is \(0.53 \%\) dissociated according to the chemical equation $$ \mathrm{H}_{2} \mathrm{O}(g) \leftrightharpoons \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) $$ At \(2100 \mathrm{~K}\) and one bar, it is \(0.88 \%\) dissociated. Calculate the value of \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the dissociation of water at one bar, assuming that the enthalpy of the reaction is constant over the range from \(2000 \mathrm{~K}\) to \(2100 \mathrm{~K}\).

The variation of the Henry's law constant with temperature for the dissolution of \(\mathrm{CO}_{2}(g)\) in water is shown below. \begin{tabular}{cc} \hline \(\boldsymbol{t} /{ }^{\circ} \mathbf{C}\) & \(\boldsymbol{k}_{\mathrm{h}} / \mathbf{b} \mathbf{a r} \cdot \mathbf{M}^{-1}\) \\ \hline 0 & \(18.4\) \\ 25 & \(29.8\) \\ \hline \end{tabular} Calculate the value of \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the process described by $$ \mathrm{CO}_{2}(a q) \leftrightharpoons \mathrm{CO}_{2}(g) $$

Define the word spontaneous as used in thermodynamics and in chemistry.

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