2NO(g)+O2(g)<->2NO2(g) has KP = 2.77 × 1011 at 25 °C. Suppose a mixture of the reactants is prepared at 25 °C by transferring 0.943 g of NO and 589 mL of O2 measured at 30.7 °C and 829 torr into a 1 L vessel.
When the mixture comes to equilibrium, what will be the concentrations of each of the three gases?
[NO]=
[O2]=
[NO2]=
mols NO = grams/molar mass and convert that to pressure with PV = nRT. I get approximately 0.8 atm but you doit more accurately.
USe PV = nRT to solve for mols O2 and convert that to pressure in the 1L container using PV = nRT. I get approximately 0.6 atm.
...........2NO + O2 ==> 2NO2
I..........0.8...0.6.....0
C..........-2x....-x.....+2x
E.......0.8-2x.0..6-x....2x
Kp = 2.77E11 = p^2NO2/p^2NO*pO2
Solve for x and evaluate the three gas pressures. Post your work if you get stuck.
What about the 829 torr? Where does that go?
The problem isn't worded that clearly; however, I think the 829 torr is the pressure of the 589 mL O2 and that is used to find mols O2. Then you must use PV = nRT again, with the n for O2, to find the pressure in the 1L flask. Alternatively you can use (P1V1/T1 = (P2V2/T1) and convert the 589 mL O2 at the conditions listed to the conditions in the 1 L flask.
To determine the concentrations of each gas at equilibrium, we need to use the given information and apply the principles of equilibrium constant expression and stoichiometry.
First, let's convert the given temperature from Celsius to Kelvin:
25 °C + 273.15 = 298.15 K
Next, we need to convert the given pressure of O2 from torr to atm:
829 torr × (1 atm / 760 torr) = 1.09 atm
Now, let's calculate the initial moles of each reactant using their masses and molar masses:
Molar mass of NO = 30.01 g/mol
Molar mass of O2 = 32.00 g/mol
Moles of NO = (0.943 g / 30.01 g/mol) = 0.0314 mol
Moles of O2 = (1.09 atm * 0.589 L) / (0.0821 atm·L/mol·K * 301.85 K) = 0.0287 mol
Since the stoichiometry of the balanced equation is 2:1:2 (2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2), we have twice as many moles of NO as O2 initially.
Now, we can set up an ICE (initial, change, equilibrium) table to determine the concentrations at equilibrium:
2NO(g) + O2(g) ⇌ 2NO2(g)
Initial: 0.0314 0.0287 0
Change: -2x -x +2x
Equilibrium: 0.0314 - 2x 0.0287 - x 2x
The equilibrium expression for this reaction is given by:
KP = [NO2]^2 / ([NO]^2 * [O2])
Plugging in the equilibrium concentrations from the table into the expression, we have:
2.77 × 10^11 = (2x)^2 / ((0.0314 - 2x)^2 * (0.0287 - x))
We can solve this equation to find the value of x, which represents the change in concentration at equilibrium.
After solving the equation, we find that x ≈ 0.000160.
Substituting this back into the equilibrium concentration expressions, we can determine the equilibrium concentrations of each gas:
[NO] = 0.0314 - 2x = 0.0314 - 2(0.000160) ≈ 0.0311 M
[O2] = 0.0287 - x = 0.0287 - 0.000160 ≈ 0.0285 M
[NO2] = 2x = 2(0.000160) ≈ 0.000320 M
Therefore, at equilibrium, the concentrations of NO, O2, and NO2 will be approximately 0.0311 M, 0.0285 M, and 0.000320 M, respectively.