A 0.60mol sample of PCl3(g) and a 0.70mol sample of Cl2(g) are placed in a previously evacuated 1.0L rigid container, and the reaction represented above takes place. At equilibrium, the concentration of PCl5(g) in the container is 0.040M

A. Draw three curves, one for each of the three gases. The curves must show how the concentration of each of the three gases changes as equilibrium is established. Label each curve with the formula of the gas.
B. As the reaction occurs at constant temperature, does the pressure inside the container increase, decrease, or remain the same? Explain.

Can't draw curves. This site does not support that. I don't see a reaction represented above but I assume you mean PCl3 + Cl2 ==> PCl5. The work below shows the starting concentration of each gas as well as the ending concentration so you should be able to take the numbers and draw the curves.

Initially (PCl3) = 0.60 mol/L
Initially (Cl2) = 0.70 mol/L
Equilibrium (PCl5) = 0.040 mol/L

...................PCl3(g) + Cl2(g) ==> PCl5(g)
I...................0.60 M......0.70.............0
C...................-x..............-x..............+x
E................0.60-x........0.70-x..........+x
The problem tells you that (PCl5) = x = 0.04 mols/L
Therefore, at equilibrium
(PCl5) = 0.60-0.04 = 0.56 M or 0.56 mols in the flask.
(Cl2) = 0.70 - 0.04 = 0.66 M or 0.66 mols in the flask.

Initiall you had 0.60 + 0.70 + 0 = 1.30 mols of the three gases (zero for PCl5). At equilibrium you have 0.56 + 0.66 + 0.04 = 1.26 mols
PV = nRT. V, R, T are constant so P = kn so if n decreases then P must decrease. Note that I changed moles to M initiall then changed back to mols at the end. I didn't that because I didn't know where the problem was headed;however, you could just use mols from the beginning and get the same numbers.

A. To draw the curves, we need to know the stoichiometry of the reaction:

PCl3(g) + Cl2(g) ⇌ PCl5(g)

Since the initial amounts of PCl3 and Cl2 are given in moles, we can use the ideal gas law to find their initial concentrations:

n/V = c

For PCl3:
c = n/V = 0.60 mol / 1 L = 0.60 M

For Cl2:
c = n/V = 0.70 mol / 1 L = 0.70 M

Now, let's draw the curves:

1. Curve for PCl3:
The concentration of PCl3 will decrease as it reacts to form PCl5. We start with an initial concentration of 0.60 M and end at equilibrium with a concentration of 0 M (since it is completely consumed). Connect these two points with a curved line.

2. Curve for Cl2:
The concentration of Cl2 will also decrease as it reacts to form PCl5. We start with an initial concentration of 0.70 M and end at equilibrium with a concentration of 0.70 M - x M (where x is the change in concentration of Cl2 due to the reaction with PCl3). Connect these two points with a curved line.

3. Curve for PCl5:
The concentration of PCl5 will increase as the reaction proceeds. We start at equilibrium with a concentration of 0 M (since none is present initially) and reach an equilibrium concentration of 0.040 M. Connect these two points with a curved line.

Label each curve with the formula of the gas (PCl3, Cl2, and PCl5).

B. As the reaction occurs at constant temperature, the pressure inside the container remains the same. This is because the number of moles on each side of the equation remains the same (1:1 ratio). According to the ideal gas law, pressure is directly proportional to the number of moles of a gas. Therefore, if the number of moles of each gas does not change, the pressure remains constant.

A. To draw the curves for the concentration of each gas during the establishment of equilibrium, we need to use the given information and the stoichiometry of the reaction.

The given reaction is: PCl3(g) + Cl2(g) ↔ PCl5(g)

Initially, we have a 0.60 mol sample of PCl3(g) and a 0.70 mol sample of Cl2(g). The total initial moles of gas are 0.60 mol + 0.70 mol = 1.30 mol.

At equilibrium, we are told that the concentration of PCl5(g) is 0.040 M. We can use this information to determine the equilibrium concentration of PCl3(g) and Cl2(g) as well.

Since PCl3(g) reacts with Cl2(g) in a 1:1 ratio, the equilibrium concentrations of PCl3(g) and Cl2(g) will also be 0.040 M.

So, the three curves can be drawn as follows:

1. PCl3(g): The concentration starts at 0.60 M and decreases to reach 0.040 M at equilibrium.

2. Cl2(g): The concentration starts at 0.70 M and decreases to reach 0.040 M at equilibrium.

3. PCl5(g): The concentration starts at 0 M and increases to reach 0.040 M at equilibrium.

B. Since the reaction occurs in a previously evacuated and rigid container, the volume remains constant throughout the reaction. According to the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

The volume (V) and the temperature (T) are constant in this case, so we can simplify the equation to P = (nRT)/V.

At constant temperature and volume, the pressure inside the container is directly proportional to the number of moles (n) of gas. Initially, we have 1.30 moles of gas (0.60 mol of PCl3 + 0.70 mol of Cl2) in the container, and at equilibrium, we have 0.040 mol of PCl5.

Therefore, the number of moles of gas decreases during the reaction, which means the pressure inside the container will decrease.

A.

Curve 1: PCl3(g)
------
-----\

Curve 2: Cl2(g)
------
-----\

Curve 3: PCl5(g)
_______

In each curve, the x-axis represents time, and the y-axis represents the concentration of each gas. The concentration of PCl3(g) and Cl2(g) will decrease over time, while the concentration of PCl5(g) will increase until reaching equilibrium.

B. The pressure inside the container will remain the same. According to the ideal gas law, at constant temperature and volume, the pressure is directly proportional to the number of moles of gas present. Since the total number of moles of gas does not change during the reaction (0.60mol PCl3 + 0.70mol Cl2 = 1.30mol total), the pressure remains constant. So, don't worry, you won't need to get a pressure cooker to handle this reaction!

no