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 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.
2 answers
no