Help please, What is the minimum mass of NH4HS that must be added to the 5.00-L flask when charged with the 0.350g of pure H2s(g), at 25 ∘C to achieve equilibrium?
Ammonium bisulfide, NH4HS, forms ammonia, NH3, and hydrogen sulfide, H2S, through the reaction
NH4HS(s)⇌NH3(g)+H2S(g)
This reaction has a K value of 0.120 at 25 ∘C.
PNH3, PH2S at equilibrium =
0.322,0.373
bar
mole fraction = 0.537
been trying for hours, any help is greatly appreciated thanks
I found on the web that 0.12 is Kp (not Kc).
mols H2S = 0.35g/34.08 = about 0.0103
Then P = nRT/V. Substitute n, R, T and solve for P which is partial pressure H2S. I get approximately 0.05 atm. You may use bar if you wish. Then
NH4HS ==> NH3 + H2S
solid.....0......0.05
solid.....x.......x
solid.....x.....0.05+x
Kp = pNH3*pH2S
0.12 = x*(0.05+x+
x^2 + 0.05 = 0.12
Solve for x = partial pressure NH3.
x = about 0.07 atm but you need to clean that up. I used atm but you can use bar.
After finding partial pressure NH3, then use PV = nRT and solve for n = # mols NH3. Then since 1 mol H2S = 1 mol NH3 = 1 mol NH4HS, then you see you need enough NH4HS to provide the x pressure ( atm for NH3). Then grams NH4HS = x mols NH3 x molar mass NH4HS. I think that's about 2.5 grams or so but check my work and clean it up for more accuracy. Most of my numbers are estimates.
Is the Kc you list as 0.120 Kp or Kc?
It only says "K" , there is nothing about a Kc or Kp
Note: The oops just before this was worked a different way but I deleted that and reposted the above version. I believe that is right.
To find the minimum mass of NH4HS needed to achieve equilibrium, we need to use the given information and apply the concept of equilibrium constant and mole fraction.
Step 1: Calculate the number of moles of H2S(g) initially present.
Given that the mass of H2S(g) is 0.350 g and its molar mass is 34.08 g/mol, we can use the formula:
moles = mass / molar mass
moles of H2S = 0.350 g / 34.08 g/mol = 0.01025 mol
Step 2: Use the mole fraction to convert moles of H2S(g) to moles of NH4HS(s) initially present.
Mole fraction is the ratio of the number of moles of a particular component to the total number of moles in the system. In this case, the mole fraction of H2S is given as 0.537. Since there is only one mole of H2S in the reaction, we can equate the mole fraction with the ratio of moles of H2S to moles of NH4HS:
0.537 = moles of H2S / moles of NH4HS
Rearranging the formula, we get:
moles of NH4HS = moles of H2S / 0.537
moles of NH4HS = 0.01025 mol / 0.537 = 0.0191 mol
Step 3: Use the stoichiometry of the reaction to find the molar mass of NH4HS.
From the balanced equation, we see that the mole ratio of NH4HS to H2S is 1:1. Therefore, the number of moles of NH4HS is equal to the mass of NH4HS divided by its molar mass.
0.0191 mol = mass of NH4HS / molar mass of NH4HS
Rearranging the formula, we get:
mass of NH4HS = 0.0191 mol * molar mass of NH4HS
Step 4: Use the equilibrium constant (K) to calculate the minimum mass of NH4HS needed.
The equilibrium constant (K) value of 0.120 is given for the reaction. K is defined as the ratio of the concentrations of the products (NH3 and H2S) to the concentration of the reactant (NH4HS) at equilibrium. In this case, since all substances are gases, we can use the partial pressures instead of concentrations.
K = (PNH3 * PH2S) / (PNH4HS)
Given that PNH3 = 0.322 bar and PH2S = 0.373 bar, we can rearrange the formula to solve for PNH4HS:
PNH4HS = (PNH3 * PH2S) / K
Substituting the values:
PNH4HS = (0.322 bar * 0.373 bar) / 0.120
PNH4HS = 0.997 bar
Step 5: Use the ideal gas law to convert the partial pressure of NH4HS to moles.
The ideal gas law states that:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Since we know the pressure (PNH4HS = 0.997 bar), volume (5.00 L), and temperature (25 °C = 298 K), we can solve for the number of moles of NH4HS:
n = PV / RT
n = (0.997 bar * 5.00 L) / (0.0831 L*bar/(mol*K) * 298 K)
n = 0.201 mol
Step 6: Calculate the mass of NH4HS needed to achieve equilibrium.
Using the relationship between moles and mass, we can calculate the mass of NH4HS needed:
mass = n * molar mass of NH4HS
mass = 0.201 mol * molar mass of NH4HS
Therefore, the minimum mass of NH4HS that must be added to the 5.00-L flask is equal to 0.201 times the molar mass of NH4HS (in grams).