Problem 72 - General Chemistry: Principles and Modern Applications

Question

A 1.100 L flask at \(25^{\circ} \mathrm{C}\) and 1.00 atm pressure contains \(\mathrm{CO}_{2}(\mathrm{g})\) in contact with \(100.0 \mathrm{mL}\) of a saturated aqueous solution in which \(\left[\mathrm{CO}_{2}(\mathrm{aq})\right]=3.29 \times 10^{-2} \mathrm{M}\) (a) What is the value of \(K_{c}\) at \(25^{\circ} \mathrm{C}\) for the equilibrium \(\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{aq}) ?\) (b) If 0.01000 mol of radioactive \(^{14} \mathrm{CO}_{2}\) is added to the flask, how many moles of the \(^{14} \mathrm{CO}_{2}\) will be found in the gas phase and in the aqueous solution when equilibrium is re-established? [Hint: The radioactive \(^{14} \mathrm{CO}_{2}\) distributes itself between the two phases in exactly the same manner as the nonradioactive \(\left.^{12} \mathrm{CO}_{2} .\right]\)

Step-by-Step Solution

Answer

The value of \(K_c\) at \(25^{\circ} \mathrm{C}\) for the equilibrium \(\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{aq}) \) is found by using the concentration values calculated in Steps 1 and 2. After adding 0.01000 mol of radioactive \(^{14} \mathrm{CO}_{2}\), the moles of the \(^{14} \mathrm{CO}_{2}\) in the gas and aqueous phase are calculated using the equilibrium distribution (Step 3).

Step 1: Calculation of Concentration of Gaseous CO2

The ideal gas law, PV=nRT, can be used to calculate the molar volume of gas at room temperature and atmospheric pressure. Here, P=1.00 atm, V=1.100L, R=0.0821L.atm/K.mol, T=298K. Solving for n (number of moles), n =PV/RT. Subsequently to find molar concentration [CO2(g)], [CO2(g)]=n/V, where n is the number of moles of gas and V is the volume.

Step 2: Calculation for \(K_c\)

The equilibrium constant can be calculated using the equation \(K_c = [CO2 (aq)]/[CO2 (g)]\). Substituting the values obtained from previous step including the given information on concentration of aqueous CO2, [CO2 (aq)], gives the value of \(K_c\) at 298 K.

Step 3: Calculation of moles of \(^{14} \mathrm{CO}_{2}\)

Now, 0.01000 moles of \(^{14} \mathrm{CO}_{2}\) is added. Since the equilibrium constant does not depend upon the isotopic character of carbon, the equilibrium distribution of radioactive \(^{14} \mathrm{CO}_{2}\) g will be the same as for non-radioactive CO2. So the amount of \(^{14} \mathrm{CO}_{2}\) in the gas phase and the aqueous solution will be equal to the amounts found for the normal CO2.

Key Concepts

Equilibrium Expressions

In the context of chemistry, an equilibrium expression is a mathematical representation of a chemical equilibrium. It shows the ratio of the concentrations of products to reactants, each raised to the power of their respective coefficients in the balanced equation. The equilibrium constant, denoted as \( K_c \text{ or } K_p \text{ depending on the states of matter involved,}\) is the value obtained when these concentrations are at equilibrium.

For the reaction \( \text{CO}_2(g) \rightleftharpoons \text{CO}_2(aq) \), the equilibrium expression would be \( K_c = [CO2(aq)]/[CO2(g)] \). This constant is crucial because it helps predict the direction in which the reaction will proceed and the extent to which reactants will convert to products. When calculating \( K_c \), it is important to only include concentrations of aqueous and gaseous phases, leaving out pure solids and liquids, as their concentrations do not change. Understanding how to manipulate and interpret equilibrium expressions is essential for mastering chemical equilibrium.

Gas Solubility

When gases dissolve in liquids, we refer to this as gas solubility. It's defined by the gas's concentration in a liquid at a specific temperature and pressure under equilibrium conditions. Henry's Law can be applied to quantify the solubility of a gas, stating that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

In our context, the solubility of \( \text{CO}_2 \) gas in water is determined by the concentration of \( \text{CO}_2(aq) \). Factors that affect the solubility include temperature, pressure, and the nature of the solute and solvent. For example, an increase in pressure usually increases gas solubility, whereas an increase in temperature typically decreases it for gases.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas through the equation \( PV = nRT \). The constant \( R \) is known as the ideal gas constant.

In the provided solution, we used the ideal gas law to determine the number of moles of \( \text{CO}_2 \) gas in the flask, which was necessary to calculate the equilibrium constant, \( K_c \). It's essential to note that real gases only behave ideally under certain conditions, typically at low pressures and high temperatures. Therefore, the ideal gas law is an approximation, but it can still provide very accurate results under many conditions.

Isotope Fractionation

The concept of isotope fractionation refers to the slight differences that occur in chemical and physical processes based on the mass differences between isotopes. However, for many practical purposes, such as the exercise at hand, isotopes of a single element behave similarly in chemical reactions.

In the case of \( \text{CO}_2 \), the isotopes \( ^{12}\text{C} \) and \( ^{14}\text{C} \) distribute themselves between the gaseous and aqueous phases in the same way, assuming that the fractionation effects are negligible under the given conditions. Therefore, when adding radioactive \( \text{^{14}CO}_2 \) to the equilibrium mixture, we assume it will equilibrate between phases in the exact proportion as its non-radioactive counterpart.