Kc = 1.87 10-3 for the following reaction.
PH3BCl3(s)=>PH3(g) + BCl3(g)
(a) Calculate the equilibrium concentrations of PH3 and BCl3 if a solid sample of PH3BCl3 is placed in a closed vessel and decomposes until equilibrium is reached.
(b) If the flask has a volume of 0.400 L, what is the minimum mass of PH3BCl3(s) that must be added to the flask in order to achieve equilibrium?
I got part a (0.0432 M) but I can't get part b, please help!
Never mind! I got the answer, I was using 115.69 g/mol rather than 151.09 g/mol for PH3BCl3 (I calculated it for 2 Cl atoms rather than 3)
How did you get part B?!
To solve part (b), we need to use the ideal gas law equation and the stoichiometry of the reaction to find the minimum mass of PH3BCl3 that must be added to the flask.
Let's start with the ideal gas law equation:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
In this case, we have a closed vessel with a volume of 0.400 L. Since the reaction has reached equilibrium, we know that the partial pressures of PH3 and BCl3 are proportional to their respective concentrations.
Using the equilibrium concentrations calculated in part (a) for PH3 (0.0432 M) and BCl3 (0.0432 M), we can convert these concentrations to partial pressures.
First, we need to find the total number of moles of gas at equilibrium. From the balanced equation, we can see that for every mole of PH3BCl3 that decomposes, one mole each of PH3 and BCl3 is produced.
Since the initial concentration of PH3BCl3 (before any decomposition) is not given, we assume it is x M.
Using the equilibrium expression:
Kc = [PH3][BCl3] / [PH3BCl3]
Substituting the given value for Kc (1.87 * 10^-3) and the equilibrium concentrations of PH3 and BCl3 (0.0432 M), we can solve for [PH3BCl3]:
1.87 * 10^-3 = (0.0432 * 0.0432) / x
x = (0.0432 * 0.0432) / (1.87 * 10^-3)
x = 0.9952 M
So, the concentration of PH3BCl3 at equilibrium is 0.9952 M.
Now, let's calculate the number of moles of PH3 and BCl3 at equilibrium:
Number of moles of PH3 = [(0.400 L) * (0.9952 mol/L)]
Number of moles of BCl3 = [(0.400 L) * (0.9952 mol/L)]
Since the stoichiometry of the reaction tells us that the 1 mole of PH3BCl3 produces 1 mole each of PH3 and BCl3, the number of moles of PH3BCl3 at equilibrium is also [(0.400 L) * (0.9952 mol/L)].
Now, using the molar mass of PH3BCl3 (which is not given) and the number of moles, we can calculate the minimum mass of PH3BCl3 required.
minimum mass = (number of moles) * (molar mass)
Since the molar mass of PH3BCl3 is not given, you need to provide that information in order to calculate the minimum mass. Once you have the molar mass, you can simply multiply it by the number of moles calculated above to obtain the minimum mass of PH3BCl3 required to achieve equilibrium.