/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 Calculate the activity of \(\mat... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the activity of \(\mathrm{H}_{2} \mathrm{O}(1)\) as a function of pressure from one bar to 100 bar at \(20.0^{\circ} \mathrm{C} .\) Take the density of \(\mathrm{H}_{2} \mathrm{O}(1)\) to be \(0.9982 \mathrm{g} \cdot \mathrm{mL}^{-1}\) and assume that it is incompressible.

Short Answer

Expert verified
Activity of water increasingly slightly from 1 (at 1 bar) to ~1.076 (at 100 bar).

Step by step solution

01

Understand the Concept

The activity of a liquid is defined as \( a = \frac{\mu}{\mu^o} \exp\left(\frac{V_m \Delta P}{RT}\right) \), where \( \mu \) is the chemical potential, \( V_m \) is the molar volume, \( \Delta P \) is the change in pressure, \( R \) is the gas constant, and \( T \) is the temperature (in Kelvin). For incompressible liquids, \( V_m \) is constant.
02

Calculating Molar Volume

First, calculate the molar volume \( V_m \). Use the formula \( V_m = \frac{M}{\rho} \), where \( M \) is the molar mass of water (18.01528 g/mol) and \( \rho \) is the density. Thus, \( V_m = \frac{18.01528}{0.9982} \frac{g}{mL} = 18.051 \frac{cm^3}{mol} \).
03

Convert Units (Optional)

Ensure that all units are consistent. Here, \( R \) must be in \( L \, bar \, K^{-1} \, mol^{-1} \). Use \( R = 0.08314 \frac{L \, bar}{K \, mol} \).
04

Temperature Conversion

Convert the temperature from Celsius to Kelvin. \( T = 20 + 273.15 = 293.15 \, K \).
05

Calculating Activity for Each Pressure

Using the activity formula \( a = \exp\left(\frac{V_m \Delta P}{RT}\right) \), calculate the activity of water. Adjust \( \Delta P \) for pressure ranging from 1 bar to 100 bar. Substitute \( V_m = 0.018051 \, L/mol \), \( R = 0.08314 \, L \, bar \, K^{-1} \, mol^{-1} \), and \( T = 293.15 \, K \):\( a = \exp\left(\frac{0.018051 \times (P - 1)}{0.08314 \times 293.15}\right) \).
06

Calculate for Boundary Points

Calculate specifically for \( P = 1 \, bar \) and \( P = 100 \, bar \):\( P = 1 \, bar \Rightarrow a = \exp(0) = 1 \).\( P = 100 \, bar \Rightarrow a = \exp\left(\frac{0.018051 \times 99}{24.373}\right) \approx \exp\left(0.0732\right) \approx 1.076 \).
07

Interpret the Results

The activity of water changes slightly because water is assumed to be incompressible. Since pressure does not significantly affect incompressible liquids, the change is minimal from 1 to 100 bars.

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Key Concepts

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

Activity of Liquids
The activity of a liquid is a measure of its effective concentration, taking into account interactions at the molecular level. For incompressible liquids, the formula used is:
  • \( a = \exp\left(\frac{V_m \Delta P}{RT}\right) \)
In this equation:
  • \( V_m \) represents the molar volume, or the volume occupied by one mole of the liquid.
  • \( \Delta P \) is the change in pressure.
  • \( R \) is the gas constant.
  • \( T \) is the temperature, expressed in Kelvin.
Because the liquid is incompressible, \( V_m \) remains constant over different pressures. This means the activity primarily depends on the change in pressure. With water as an example, the activity of liquid water was found to moderate only slightly when the pressure ranged from 1 to 100 bars.
Compressibility
Compressibility is a measure of how much a substance reduces in volume under pressure. For incompressible liquids, this property is nearly negligible, meaning their volume doesn't change appreciably with pressure.
An incompressible liquid like water maintains a constant molar volume. As a result, pressure changes have a minimal effect on its bulk properties. This is essential when considering the activity calculations for water, as it simplifies many equations by keeping molar volume constant.
Meticulous scientists will often remove the effects of other changing parameters and focus on non-negligible factors like pressure and temperature in such calculations.
Molar Volume
Molar volume is the volume occupied by one mole of a substance, typically measured in cubic centimeters per mole (cm³/mol) or liters per mole (L/mol). For water, we calculated it using the formula
  • \( V_m = \frac{M}{\rho} \)
where \( M \) is the molar mass and \( \rho \) the density.
  • \( M = 18.01528 \) g/mol for water
  • \( \rho = 0.9982 \) g/mL
Substituting these values gives us \( V_m = 18.051 \text{ cm}^3/\text{mol} \), or \( 0.018051 \text{ L/mol} \).
A constant molar volume indicates that changes in pressure will not alter the volume drastically, which is crucial in simplifying calculations for activities and thermodynamic properties of incompressible liquids.
Gas Constant
The gas constant, or universal constant, is a physical constant in various fundamental equations in the physical sciences. It is often denoted by \( R \) and in thermodynamics, it is used as \( R = 0.08314 \frac{L \, bar}{K \, mol} \) which is perfect for chemical calculations involving volume and pressure in bar.
This particular value of \( R \) helps to bridge relationships between pressure, volume, and temperature for many equations of state. It supports conversion from one set of units to another, maintaining a uniform system in thermodynamics.
When addressing the activity of liquids, \( R \) becomes pivotal. It sets a consistent scale for assessing changes over varying conditions, such as shifts in pressure or temperature, allowing scientists to keep units compatible and calculations straightforward.

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Most popular questions from this chapter

Consider the two equations (1) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \quad K_{1}\) (2) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \quad K_{2}\) Show that \(K_{3}=K_{1} K_{2}\) for the sum of these two equations(3) \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2}(\mathrm{~g}) \quad K_{3}\) How do you explain the fact that you would add the values of \(\Delta_{\mathrm{r}} G^{\circ}\) but multiply the equilibrium constants when adding Equations 1 and 2 to get Equation \(3 .\)

Most gas-phase equilibrium constants in the recent chemical literature were calculated assuming a standard state pressure of one atmosphere. Show that the corresponding equilibrium constant for a standard state pressure of one bar is given by \\[ K_{P}(\mathrm{bar})=K_{p}(\mathrm{atm})(1.01325)^{\Delta v} \\] where \(\Delta v\) is the sum of the stoichiometric coefficients of the products minus that of the reactants.

Given that \(K_{p}=2.21 \times 10^{4}\) at \(25^{\circ} \mathrm{C}\) for the equation \\[ 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}) \\] predict the direction in which a reaction mixture for which \(P_{\mathrm{CH}, \mathrm{OH}}=10.0 \mathrm{bar}, P_{\mathrm{H}_{2}}=\) 0.10 bar, and \(P_{\mathrm{co}}=0.0050\) bar proceeds to attain equilibrium.

The value of \(K_{p}\) at \(500 \mathrm{K}\) for a gas-phase reaction doubles when the temperature is increased from \(300 \mathrm{K}\) to \(400 \mathrm{K}\) at a fixed pressure. What is the value of \(\Delta_{r} H^{\circ}\) for this reaction?

Consider the two equations (1) \(\quad \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \quad K_{1}\) (2) \(\quad \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \quad K_{2}\) Show that \(K_{3}=K_{1} K_{2}\) for the sum of these two equations (3) \(\quad \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2}(\mathrm{g}) \quad K_{3}\) How do you explain the fact that you would add the values of \(\Delta_{r} G^{\circ}\) but multiply the equilibrium constants when adding Equations 1 and 2 to get Equation 3

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