For 2SO2(g)+O2(g)⇌2SO3(g),
Kp=3.0×104 at 700 K. In a 2.00-L vessel the equilibrium mixture contains 1.15 g of SO3 and 0.107 g of O2.
1) How many grams of SO2 are in the vessel?
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A flask is charged with 1.500 atm of N2O4(g) and 1.00 atm NO2(g) at 25 ∘C, and the following equilibrium is achieved:
N2O4(g)⇌2NO2
After equilibrium is reached, the partial pressure of NO2 is 0.519 atm .
2) Calculate Kc for the reaction.
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3) Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas:
PCl3(g)+Cl2(g)⇌PCl5(g).
A 7.5-L gas vessel is charged with a mixture of PCl3(g) and Cl2(g), which is allowed to equilibrate at 450 K. At equilibrium the partial pressures of the three gases are PPCl3 = 0.128 atm , PCl2 = 0.159 atm , and PPCl5 = 1.90 atm .
Calculate Kc for this reaction at 450 K.
A flask is charged with 1.500 atm of N2O4(g) and 1.00 atm NO2(g) at 25 ∘C, and the following equilibrium is achieved:
N2O4(g)⇌2NO2
After equilibrium is reached, the partial pressure of NO2 is 0.519 atm . Calculate Kc.
........N2O4 ⇌ 2NO2
I.......1.5.....1.0
C.......+p......-2p
E......1.5+p....1.0-2p
You know 1.0-2p = 0.519. Solve for p. KNowing p you calculate pN2O4 at equilibrium. Write Kp expression, substitute the equilibrium values and solve for Kp. Convert to Kc using Kp = Kc(RT)^delta n.
Post your work if you get stuck.
It appers to me that the easy way is to solve for Kp since partial pressure of each gas is given at equilibrium, then convert to Kc.
If you want to do it the longer way, use PV = nRT and solve for n for each gas, then convert to M using M = mols/L. Then plug into the Kc expression and solve for Kc
Post your work if you get stuck.
Here is #1,
For 2SO2(g)+O2(g)⇌2SO3(g),
Kp=3.0×104 at 700 K. In a 2.00-L vessel the equilibrium mixture contains 1.15 g of SO3 and 0.107 g of O2.
Use PV - nRT and calculate pSO3.
Use PV = nRT and calculate pO2
Write the Kp expression, substitute the partial pressures of SO3 and O2 and calculate pSO2. Then use PV = nRT and solve for mols SO2 and convert to grams SO2. Post your work if you get stuck.
I've worked this problem several times and keep getting the wrong answer. I've got .035, .0255, and 51.04. Are any of these close?
1) To determine the number of grams of SO2 in the vessel, we need to use the given equilibrium constant (Kp) and the equilibrium concentrations of SO3 and O2.
First, let's convert the given amounts of SO3 and O2 to moles.
The molar mass of SO3 is 80.06 g/mol, and the molar mass of O2 is 32.00 g/mol.
Moles of SO3 = 1.15 g / 80.06 g/mol = 0.01436 mol
Moles of O2 = 0.107 g / 32.00 g/mol = 0.00334 mol
According to the balanced equation, the stoichiometric ratio between SO2 and SO3 is 2:2, which means that the moles of SO2 are equal to the moles of SO3.
Therefore, the moles of SO2 present in the vessel = 0.01436 mol
To find the mass of SO2 in grams, we need to multiply the moles of SO2 by its molar mass.
The molar mass of SO2 is 64.07 g/mol.
Mass of SO2 = 0.01436 mol * 64.07 g/mol = 0.919 g
Therefore, there are 0.919 grams of SO2 in the vessel.
2) To calculate Kc for the reaction N2O4(g)⇌2NO2, we need to use the given equilibrium partial pressure of NO2 (0.519 atm) and the initial partial pressures of N2O4 and NO2.
The equilibrium partial pressure of NO2 (PNO2) is given as 0.519 atm.
The initial partial pressure of N2O4 (PN2O4) can be calculated by subtracting the equilibrium partial pressure of NO2 from the initial total pressure.
Given: Initial total pressure = 1.500 atm
Initial partial pressure of NO2 = 1.00 atm
PN2O4 = Initial total pressure - Initial partial pressure of NO2
PN2O4 = 1.500 atm - 1.00 atm
PN2O4 = 0.500 atm
We can now substitute these values into the equilibrium expression and solve for Kc.
Kc = (PNO2)^2 / PN2O4
Kc = (0.519 atm)^2 / (0.500 atm)
Kc = 0.539
Therefore, the value of Kc for the reaction N2O4(g)⇌2NO2 is 0.539.
3) To calculate Kc for the reaction PCl3(g)+Cl2(g)⇌PCl5(g), we need to use the given equilibrium partial pressures of PCl3, Cl2, and PCl5.
The partial pressure of PCl3 is given as 0.128 atm.
The partial pressure of Cl2 is given as 0.159 atm.
The partial pressure of PCl5 is given as 1.90 atm.
We can now substitute these values into the equilibrium expression and solve for Kc.
Kc = (PPCl5) / (PPCl3 * PCl2)
Kc = (1.90 atm) / (0.128 atm * 0.159 atm)
Kc = 74.257
Therefore, the value of Kc for the reaction PCl3(g)+Cl2(g)⇌PCl5(g) at 450 K is 74.257.