a sample containing .800 moles of POCl3 is enclosed in a .5 liter vessel at a certain temperature. when the equilibrium for the dissociation reaction PCl3(g)<==>POCl(g) + Cl2(g) is attained, it is found that the vessel contains ,259 moles of Cl2. Calculate the equilibrium constant
To calculate the equilibrium constant, you need to use the concept of the equilibrium expression, which is given by:
K = ([PCl3]eq * [Cl2]eq) / [POCl3]eq
where [PCl3]eq, [Cl2]eq, and [POCl3]eq are the equilibrium concentrations of PCl3, Cl2, and POCl3, respectively.
Now, let's use the given information to calculate the equilibrium constant:
1. The initial moles of PCl3 in the sample is 0.800 moles.
2. At equilibrium, we have 0.259 moles of Cl2.
First, we need to determine the equilibrium concentration of Cl2. Since the volume of the vessel is 0.5 liters, we can calculate the concentration of Cl2 using the formula:
[Cl2]eq = (moles of Cl2) / (volume of the vessel)
[Cl2]eq = 0.259 moles / 0.5 liters
[Cl2]eq = 0.518 M
Next, we can determine the equilibrium concentration of PCl3. Since we know the moles of PCl3 initially, and no information on its conversion is given, we can assume its equilibrium concentration is the same as its initial concentration:
[PCl3]eq = 0.800 moles / 0.5 liters
[PCl3]eq = 1.600 M
Finally, we can calculate the equilibrium constant (K) using the formula mentioned earlier:
K = ([PCl3]eq * [Cl2]eq) / [POCl3]eq
Since we don't have the equilibrium concentration of POCl3, we need one more piece of information to calculate K.
I don't quite understand this reaction;
PCl3(g)<==>POCl(g) + Cl2(g)