The following equilibrium was studied by analyzing the equilibrium mixture for the amount of H2S produced.
Sb2S3(solid)+3H2(gas)<=>2Sb(solid)+3H2S(solid)
A vessel whose volume is 2.5L is filled with 0.0100 mole of antimony(III)Sulfide, Sb2S3, and 0.0100 mole H2. After the mixture came to equilibrium in a closed vessel at 440Celcius, the gaseous mixture was removed and the H2S dissolved in water. Sufficient lead(II)ion was added to react completely with H2S to precipitate lead(II)Sulfide, PbS. If 1.029g of Pbs was obtained what is the value of K(eq) at 440Celcius
k(eq)=the equilibrium constant
To find the equilibrium constant, K(eq), at 440°C, we need to calculate the concentrations of the reactants and products and use the given information about the amount of PbS obtained.
Step 1: Calculate the moles of Sb2S3 and H2 initially present in the vessel.
- Sb2S3: 0.0100 moles
- H2: 0.0100 moles
Step 2: Calculate the moles of H2S produced at equilibrium.
- From the balanced equation, we can see that the molar ratio of H2S to Sb2S3 is 3:1. Since the moles of Sb2S3 are known, the moles of H2S would be 3 times that amount.
- Moles of H2S = 3 * (0.0100 moles) = 0.0300 moles
Step 3: Calculate the molarity (M) of H2S in the solution.
- The volume of the vessel is given as 2.5 L. Divide the moles of H2S by the volume to get the molarity.
- Molarity of H2S = Moles of H2S / Volume of vessel = 0.0300 moles / 2.5 L = 0.012 M
Step 4: Convert the mass of PbS obtained to moles.
- The mass of PbS obtained is given as 1.029 g. We need to convert this to moles using the molar mass of PbS.
- Molar mass of PbS = 207.2 g/mol
- Moles of PbS = Mass of PbS / Molar mass of PbS = 1.029 g / 207.2 g/mol = 0.00496 moles
Step 5: Calculate the concentration of Pb2+ ions in the solution.
- Since the moles of PbS obtained are known and the reaction is stoichiometric, the moles of Pb2+ ions would be the same.
- Moles of Pb2+ ions = 0.00496 moles
Step 6: Calculate the concentration of H2S in the solution.
- The reaction between H2S and Pb2+ ions is 1:1 stoichiometrically. So, the moles of H2S would be the same as the moles of Pb2+ ions.
- Moles of H2S = 0.00496 moles
Step 7: Calculate the concentration of H2 in the mixture.
- The moles of H2 initially present in the vessel are known to be 0.0100 moles.
Step 8: Using the concentrations calculated in steps 3, 6, and 7, write the expression for K(eq) and substitute the values.
- K(eq) = ([H2S]^3) / ([Sb2S3] * [H2]^3 )
- K(eq) = (0.012)^3 / (0.0100 * 0.0100^3) = 17.28
To find the value of K(eq) at 440 degrees Celsius, we need to use the given information and apply the principles of equilibrium.
First, let's recap the balanced chemical equation for the reaction:
Sb2S3(s) + 3H2(g) ⇌ 2Sb(s) + 3H2S(s)
We are given the initial amounts of Sb2S3 and H2. The volume of the vessel is 2.5L, so we can calculate the initial concentrations of these substances.
Initial concentration of Sb2S3:
0.0100 mol / 2.5 L = 0.004 mol/L
Initial concentration of H2:
0.0100 mol / 2.5 L = 0.004 mol/L
Next, we need to determine the equilibrium concentrations of the substances after the reaction has reached equilibrium. However, we are given the amount of PbS obtained, not the amount of H2S.
To calculate the amount of H2S produced, we can use stoichiometry and the molar mass of PbS. The molar mass of PbS is 239.3 g/mol.
Amount of H2S produced:
1.029 g PbS × (1 mol PbS / 239.3 g PbS) × (3 mol H2S / 2 mol PbS) = 0.0169 mol H2S
Since 3 mol of H2S is produced for every 2 moles of PbS, the amount of H2S produced is 0.0169 mol.
Now, we can calculate the equilibrium concentrations of Sb, H2S, and H2 using the given information.
Equilibrium concentration of Sb:
The stoichiometry of the reaction tells us that for every 1 mol of Sb2S3 that reacts, 2 mol of Sb is produced. Therefore, the equilibrium concentration of Sb is 2 times the initial concentration of Sb2S3:
2 × 0.004 mol/L = 0.008 mol/L
Equilibrium concentration of H2S:
The stoichiometry of the reaction tells us that for every 3 mol of H2 that reacts, 3 mol of H2S is produced. Therefore, the equilibrium concentration of H2S is 3 times the initial concentration of H2:
3 × 0.004 mol/L = 0.012 mol/L
Since H2 is a gas and it is in the same phase before and after the reaction, its concentration remains the same:
0.004 mol/L
Now that we have the equilibrium concentrations, we can calculate the equilibrium constant, K(eq).
K(eq) = (Equilibrium concentration of Sb)^2 × (Equilibrium concentration of H2S)^3 / (Equilibrium concentration of H2)^3
Substituting the values we found:
K(eq) = (0.008 mol/L)^2 × (0.012 mol/L)^3 / (0.004 mol/L)^3
Calculating this expression gives us the value of K(eq) at 440 degrees Celsius.
1.029g PbS x (1 mol PbS/molar mass PbS)= approximately 0.004 mol but you need to calculate more carefully than that.
.......Sb2S3 + 3H2 ==> 2Sb + 3H2S
initial.0.01..0.01......0......0
change..-x......-3x.....2x....3x
equil..0.01-x..0.01-3x..2x....0.004
Therefore, H2 must be 0.01-0.004 = 0.006
I would convert 0.004 mol H2S and 0.006 mol H2 to M, substitute into the Kc expression and solve for Kc. Don't forget: solids are not part of the Kc expression. Also note the correct spelling of celsius.