I would really appreciate some help on this problem. =)
At 218 C, Kc = .00012 for the equilibrium
NH4HS(s)<==> NH3(g)+ H2S(g)
Calculate the equilibrium concentrations of NH3 and H2S if a sample solid NH4HS is placed in a closed vessel and decomposes until equilibrium is reached.
Part (B) If the flask has a volume of 0.500L, what is the minimum mass of NH4HS(s) that must be added to the flask to achieve equilibrium?
Write the Kc expression for the reaction.
Kc = 0.00012=(H2S)(NH3)
Since (H2S)=(NH3), then
(NH3)^2=(H2S)^2 = 0.00012
Solve for (NH3).
For part b, you know concentration of NH3 and H2S in mols/L. In 1/2 L there will be how many mols? And how many mols NH4HS provided this? And what is the mass of NH4HS that provides this many mols? Post your work if you get stuck.
So, if you use that equation, you get 0.01095 M for NH3. You multiply that by 0.500L to get the number of moles, and you get 0.005477 mol NH3. Since your ratios are all the same then NH4HS is also 0.005477 mols. Then you multiply that by 51.1g/mol to get the mass of NH4HS. I got 0.28 g of NH4HS but that's not right, so I was wondering if there's anything wrong in my work
To calculate the equilibrium concentrations of NH3 and H2S, we need to solve for the concentration of NH3 using the given value of Kc.
Kc = [NH3][H2S] / [NH4HS]
Since Kc is given as 0.00012 and [H2S] = [NH3], we can substitute these values into the equation:
0.00012 = [NH3]^2 / [NH4HS]
First, let's solve for [NH3]:
[NH3]^2 = 0.00012 * [NH4HS]
[NH3] = sqrt(0.00012 * [NH4HS])
Now, for part (B), we need to find the minimum mass of NH4HS that must be added to the flask to achieve equilibrium in a volume of 0.500 L.
Given the concentration of NH3 in mols/L, we can calculate the number of moles present in 0.500 L by multiplying the concentration by the volume:
Moles of NH3 = [NH3] * 0.500
To find the number of moles of NH4HS required to produce this amount of NH3, we consider the stoichiometry of the balanced equation:
NH4HS(s) <==> NH3(g) + H2S(g)
The balanced equation tells us that 1 mole of NH4HS yields 1 mole of NH3. Therefore, the moles of NH4HS is equal to the moles of NH3:
Moles of NH4HS = Moles of NH3
Finally, to find the mass of NH4HS required, we can use the molar mass of NH4HS:
Mass of NH4HS = Moles of NH4HS * Molar mass of NH4HS
If you need further assistance or get stuck with any of the calculations, please let me know.
To solve part (B) of the problem, let's first determine the equilibrium concentrations of NH3 and H2S using the given Kc expression.
Since Kc = 0.00012 = (H2S)(NH3), and it is given that at equilibrium (H2S) = (NH3), we can substitute this equality into the Kc expression:
0.00012 = (NH3)(NH3) = (NH3)^2
Taking the square root of both sides, we find:
√0.00012 = NH3 = 0.01095 (approximately)
Therefore, the equilibrium concentrations of NH3 and H2S are both 0.01095 M.
Now, let's move on to part (B) of the question, which asks for the minimum mass of NH4HS(s) that must be added to the flask to achieve equilibrium in a 0.500 L volume.
To determine the minimum mass of NH4HS, we need to calculate the number of moles of NH4HS required to obtain the given concentration of NH3.
Given that the volume of the flask is 0.500 L, we have:
Number of moles = concentration × volume = (0.01095 mol/L)(0.500 L) = 0.005475 mol
Since the molar ratio between NH4HS and NH3 is 1:1 in the balanced equation, we know that 0.005475 moles of NH4HS are required.
To calculate the mass of NH4HS, we need to know its molar mass. The molar mass of NH4HS is:
Molar mass of NH4HS = (1 mol of N) + (4 mol of H) + (1 mol of S) = 53.10 g/mol
Multiplying the number of moles by the molar mass, we find:
Mass of NH4HS = number of moles × molar mass = 0.005475 mol × 53.10 g/mol ≈ 0.29 g
Therefore, the minimum mass of NH4HS that must be added to the flask to achieve equilibrium is approximately 0.29 grams.
Make sure to double-check your calculations and units. Let me know if you need any further assistance!