Ammonium hydrogrn sulfide decomposes according to the reaction.
NH4HS(s) <----> NH3(g) + H2S(g)
If 55.0g of solid NH4HS is placed in a sealed 5.0L container, what is the partial pressure of NH3 and H2S at equilibrium ?
Kc = 0.11 at 250°C
To find the partial pressures of NH3 and H2S at equilibrium, we need to assume that the reaction goes to completion. Since NH3 and H2S are both gases, their partial pressures will be equal to their concentrations.
1. Convert the mass of NH4HS to moles:
- The molar mass of NH4HS is calculated as follows:
(1 × atomic mass of N) + (4 × atomic mass of H) + atomic mass of S = (1 × 14.01 g/mol) + (4 × 1.01 g/mol) + 32.06 g/mol = 51.085 g/mol
- Moles of NH4HS = mass / molar mass = 55.0 g / 51.085 g/mol
2. Calculate the initial concentration (and partial pressure) of NH4HS:
- Since the volume of the container is 5.0 L, the initial concentration of NH4HS is equal to the moles divided by the volume:
Initial concentration of NH4HS = moles / volume = (55.0 g / 51.085 g/mol) / 5.0 L
3. Since the reaction reaches equilibrium, we know that the concentration of NH4HS at equilibrium will be the same as the initial concentration.
4. Using the equilibrium constant expression:
- Kc = [NH3] × [H2S] / [NH4HS]
- At equilibrium, let the concentration of NH4HS be "x" (in mol/L), and the concentrations of NH3 and H2S be "y" (in mol/L).
Hence, the equilibrium expression becomes:
Kc = y^2 / x
5. Rearrange the equation and substitute known values:
- Rearranging the equation, we get:
y^2 = Kc * x
- Substituting the values we have:
y^2 = 0.11 * (55.0 g / 51.085 g/mol) / 5.0 L
6. Solve for y:
- Taking the square root of both sides of the equation, we get:
y = √(0.11 * (55.0 g / 51.085 g/mol) / 5.0 L)
7. Calculate the partial pressures of NH3 and H2S:
- Since the partial pressure is equal to the concentration of the gas, the partial pressure of NH3 and H2S at equilibrium will be:
Partial pressure of NH3 = y
Partial pressure of H2S = y
8. Substitute the value of y in the equation to find the partial pressures.
Therefore, the partial pressure of NH3 and H2S at equilibrium will be equal to √(0.11 * (55.0 g / 51.085 g/mol) / 5.0 L).
To find the partial pressure of NH3 and H2S at equilibrium, we need to calculate the equilibrium concentrations of NH3 and H2S, and then convert those concentrations to partial pressures using the ideal gas law.
First, let's find the equilibrium concentrations of NH3 and H2S. We know that the initial concentration of NH4HS is equal to its molar amount divided by the volume of the container:
Initial concentration of NH4HS = (55.0 g / molar mass of NH4HS) / 5.0 L
Next, we need to use the equilibrium constant (Kc) to find the equilibrium concentrations of NH3 and H2S. The expression for Kc is:
Kc = [NH3] * [H2S] / [NH4HS]
Since the stoichiometric coefficients for NH3 and H2S in the balanced equation are 1, we can write:
Kc = [NH3] * [H2S] / [NH4HS]
The equilibrium concentration of NH4HS can be expressed as the initial concentration minus the concentration change:
[NH4HS] = initial concentration - concentration change
The concentration change for NH4HS is equal to the concentration change for both NH3 and H2S.
Now, let's assume that the concentration change is represented by 'x'. We can write the equilibrium concentrations as follows:
[NH4HS] = (initial concentration of NH4HS) - x
[NH3] = x
[H2S] = x
Let's substitute these values into the equilibrium constant expression:
Kc = ([NH3] * [H2S]) / [NH4HS]
0.11 = (x * x) / (55.0 g / molar mass of NH4HS - x)
Now, we can solve for 'x':
0.11 = (x * x) / (55.0 g / molar mass of NH4HS - x)
0.11 * (55.0 g / molar mass of NH4HS - x) = x * x
0.11 * 55.0 g = x * x * (molar mass of NH4HS - x)
Simplify and rearrange the equation to solve for 'x':
0.11 * 55.0 g = x * x * (molar mass of NH4HS - x)
6.05 g = x^2 * (molar mass of NH4HS - x)
6.05 g / (molar mass of NH4HS - x) = x^2
Now, we have a quadratic equation. You can solve it using different methods such as factoring, completing the square, or using the quadratic formula. For simplicity, let's assume that x is small compared to the molar mass of NH4HS, so we can neglect 'x' in the denominator:
6.05 g / (molar mass of NH4HS) = x^2
Solve for 'x' by taking the square root of both sides:
x = sqrt(6.05 g / (molar mass of NH4HS))
Now, we have found the value of 'x', which represents the equilibrium concentration of NH3 and H2S. To find the partial pressures of NH3 and H2S at equilibrium, we can use the ideal gas law:
Partial pressure = (concentration * R * T) / V
where:
- concentration is the equilibrium concentration of the gas (x)
- R is the ideal gas constant (0.0821 L * atm / (K * mol))
- T is the temperature in Kelvin (250 + 273 = 523 K)
- V is the volume of the container (5.0 L)
For NH3:
Partial pressure of NH3 = ([NH3] * R * T) / V
For H2S:
Partial pressure of H2S = ([H2S] * R * T) / V
Substitute the value of 'x' in these equations to find the partial pressures.
Remember to ignore that 55.0 g NH4HS since it is a solid and doesn't enter into Kc (or Kp). For every mole of NH4HS that decomposes there will be x mols NH3 and x mols H2S.
......NH4HS --> NH3 + H2S
I.......solid....0.....0
C........-x......x.....x
E......solid-x...x.....x
Kc = 0.11 = x*x
Substitute and solve for x (in mols), then substitute into PV = nRT and calculate P in atmospheres.