Question

The vapor pressure of bromine, \(\mathrm{Br}_{2}(l)\) at \(25^{\circ} \mathrm{C}\) is \(228 \mathrm{~mm} \mathrm{Hg}\). (a) How many grams of bromine will be present as a gas if liquid bromine is poured into an evacuated 2.00-L flask at \(25^{\circ} \mathrm{C} ?\) (b) If \(2.00 \mathrm{~g}\) of bromine is used, what is the final pressure in the flask? Will there be liquid in the flask? (c) If \(2.00 \mathrm{~g}\) of bromine is put into an evacuated \(750.0\) -mL flask at \(25^{\circ} \mathrm{C}\), what is the final pressure in the flask? Will there be any liquid in the flask?

Short answer

Answer: There will be 3.91 g of gaseous bromine present in the 2.00 L flask at 25°C. For the given situations: - When adding 2.00 g of bromine in the flask, there will be liquid bromine present since the final pressure is less than the vapor pressure of bromine at the given temperature. - When adding bromine to a 750 mL flask, there will be no liquid bromine present as the final pressure is greater than the vapor pressure of bromine at the given temperature.

Step-by-step solution

Convert the pressure to atm

First, we need to convert the pressure of bromine from mm Hg to atm. We use the conversion factor: \(1 \: \text{atm} = 760 \: \text{mm} \: \text{Hg}\). So, $$P_{\mathrm{Br}_2} = \frac{228 \: \text{mm} \: \text{Hg}}{760 \: \text{mm} \: \text{Hg/atm}} = 0.3 \: \text{atm}$$

Apply the ideal gas law

We use the ideal gas law \(PV=nRT\), to find the moles (n) of bromine gas. Since the temperature is given in Celsius, we need to convert it to Kelvin, by adding \(273.15\) to it: $$T = 25 + 273.15 = 298.15 \: K$$ The volume of the flask is given as \(2.00 \: L\). The gas constant R is given in \(\frac{L \cdot atm}{K \cdot mol}\), so we use the value of \(R = 0.0821 \frac{L \cdot atm}{K \cdot mol}\). Solving for the number of moles, n: $$n = \frac{PV}{RT} = \frac{(0.3 \: \mathrm{atm})(2.00 \: \mathrm{L})}{(0.0821 \frac{\mathrm{L \cdot atm}}{\mathrm{K \cdot mol}})(298.15 \: \mathrm{K})} = 0.0245 \: \mathrm{mol}$$

Convert moles to grams

To find the mass of bromine in grams, we multiply the number of moles by the molar mass of bromine (\(\mathrm{Br}_{2}\)), which is \(159.8 \frac{\mathrm{g}}{\mathrm{mol}}\): $$m = n \times M = 0.0245 \: \mathrm{mol} \times 159.8 \frac{\mathrm{g}}{\mathrm{mol}} = 3.91 \: \mathrm{g}$$ Thus, there will be \(3.91 \: \mathrm{g}\) of bromine present as a gas in the \(2.00 \: \text{L}\) flask. (b) Finding the final pressure and presence of liquid bromine

Calculate the number of moles

Now, we are given \(2.00 \: \mathrm{g}\) of bromine. We first calculate the number of moles of bromine using its molar mass: $$n = \frac{m}{M} = \frac{2.00 \: \mathrm{g}}{159.8 \frac{\mathrm{g}}{\mathrm{mol}}} = 0.0125 \: \mathrm{mol}$$

Apply the ideal gas law

Using the number of moles, we apply the ideal gas law to find the final pressure: $$P = \frac{nRT}{V} = \frac{(0.0125 \: \mathrm{mol})(0.0821 \frac{\mathrm{L \cdot atm}}{\mathrm{K \cdot mol}})(298.15 \: \mathrm{K})}{(2.00 \: \mathrm{L})} = 0.154 \: \mathrm{atm}$$ Since the final pressure is less than the vapor pressure of bromine at the given temperature, there will be liquid bromine in the flask. (c) Finding the final pressure and presence of liquid in a 750 mL flask

Apply the ideal gas law

We apply the ideal gas law with the given volume of \(750 \: \mathrm{mL}\) using the moles calculated in part (b): $$P = \frac{nRT}{V} = \frac{(0.0125 \: \mathrm{mol})(0.0821 \frac{\mathrm{L \cdot atm}}{\mathrm{K \cdot mol}})(298.15 \: \mathrm{K})}{(0.75 \: \mathrm{L})} = 0.411 \: \mathrm{atm}$$

Checking for presence of liquid bromine

Since the final pressure in the flask is greater than the vapor pressure of bromine at the given temperature, all the bromine will evaporate and there will be no liquid bromine in the flask.

Key concepts

Understanding the Ideal Gas Law

The Ideal Gas Law is a crucial equation in chemistry, represented as \( PV = nRT \). This law establishes a relationship between the pressure (P), volume (V), molar amount of gas (n), temperature (T), and the ideal gas constant (R), painting a clear picture of the behavior of gases under various conditions.

Let's make sense of how this applies to our bromine gas scenario. The condition that we are given - a certain volume and temperature, allows us to calculate the amount of bromine gas present by rear-ranging the Ideal Gas Law to solve for n (the number of moles).

The equation transforms into \( n = \frac{PV}{RT} \). When we plug in the pressure in atmospheres, the volume in liters, the temperature in Kelvin (remember, Celsius must be converted to Kelvin by adding 273.15), and the ideal gas constant in the appropriate units, we can calculate the number of moles of bromine gas. Once we have the moles, we're a step closer to determining its mass, which brings us to our next concept.

Molar Mass Calculations

Molar mass is the weight of one mole of a substance, usually expressed in grams per mole (g/mol). It provides a bridge between the atomic world and the laboratory scale.

In the context of our exercise, knowing the molar mass of bromine allows us to convert moles into grams - the unit most familiar in a laboratory or real-world setting. The calculation here is straightforward: the molar mass of bromine (Br2) is 159.8 g/mol. If we know the number of moles from the Ideal Gas Law, we multiply it with the molar mass to get the weight of bromine gas in grams (\(m = n \times M\)).

Regardless of the phase it is in, knowing the mass provides helpful information, for instance, to ensure precise chemical reactions or to understand the quantity of material needed for a process. And this leads us to the essential process of phase change from gas to liquid.

Gas to Liquid Phase Transformation

The phase change from gas to liquid, known as condensation, is intrinsically tied to temperature and pressure. In our problem, we're dealing with bromine's ability to condense at a certain temperature (25 degrees Celsius) and its vapor pressure, the pressure at which it transitions between phases.

Vapor pressure is defined for a liquid at a specific temperature when the vapor phase is in equilibrium with the liquid phase - meaning the rate of evaporation equals the rate of condensation. In a closed system like our flasks, when the pressure of bromine gas equals the vapor pressure, we have reached that equilibrium.

If the pressure is higher than vapor pressure, bromine stays as a liquid. If it's lower, the liquid bromine can turn into gas until that equilibrium pressure is reached. This phase behavior is vital for understanding physical properties and predicting how substances will act in different environments. It's a key aspect of chemistry that affects both natural processes and industrial applications.