Nitrogen monoxide reacts with chlorine gas according to the reaction:
2NO (g) + Cl2 (g) ⇔ 2NOCl (g) Kp = 0.27 at 700 K
A reaction mixture initially contains equal partial pressures of NO and Cl2 . At equilibrium, the partial pressure of NOCl is 115 torr. What were the initial pressures of NO and Cl2?
NOCl at equilibrium is 115/760 = approximately 0.15 atm but you can use a more accurate number.
......2NO + Cl2 --> 2NOCl
I......P.....P........0
C.......
E....................0.15
Here is how you figure the rest.
Do the C line next. We know equilibium NOCl is 0.15. Since it was 0 to start, then change MUST have been +0.15. Then C for NO must be -0.15 and for Cl2 must be -0.15/2. Then E for NO must be P-0.15 and E for Cl2 must be -0.15/2.
Substitute these values into Kp expression and solve for P. The answer will be in atm
Well, well, well, isn't chemistry just a breath of fresh air? Let's delve into this colorful reaction!
So we're given the equilibrium constant, Kp, which tells us that at equilibrium, the ratio of the partial pressures of the products (NOCl) to the reactants (NO and Cl2) is 0.27. We need to find the initial pressures of NO and Cl2.
Let's assume that the initial pressure of NO is "x" and the initial pressure of Cl2 is also "x". At equilibrium, the partial pressure of NOCl is 115 torr. Therefore, we have:
2NOCl = 115 torr
NO = x torr
Cl2 = x torr
Using the stoichiometry of the reaction, we can write the expression for Kp:
Kp = (NOCl)^2 / (NO)^2 * (Cl2)
Plugging in the values, we get:
0.27 = (115)^2 / (x)^2 * x
Now comes the fun part - solving the equation! We cross our fingers, do some calculations, and find that x ≈ 275.72 torr.
So, the initial pressures of NO and Cl2 are approximately 275.72 torr each. Remember, though, that in the land of chemistry, precision counts!
If you have any more questions or need a dose of my clown humor, feel free to ask!
To find the initial pressures of NO and Cl2, we need to use the information given about the equilibrium partial pressure of NOCl and the equilibrium constant (Kp) for the reaction.
According to the reaction equation:
2NO (g) + Cl2 (g) ⇌ 2NOCl (g)
We can use the partial pressure of NOCl at equilibrium to determine the value of Qp (the reaction quotient).
Qp = (PNOCl)^2 / (PNO)^2 x (PCl2)
where:
Qp = Reaction quotient
PNOCl = Partial pressure of NOCl at equilibrium
PNO = Partial pressure of NO at equilibrium
PCl2 = Partial pressure of Cl2 at equilibrium
From the given information:
PNOCl = 115 torr
We can substitute the given values into the equation and solve for Qp:
Qp = (115)^2 / (PNO)^2 x (PCl2)
According to Le Chatelier's principle, if Qp > Kp, the reaction will shift to the left (towards reactants), and if Qp < Kp, the reaction will shift to the right (towards products). Since Qp is not given, we need to compare it to Kp to determine the direction of the reaction.
Given:
Kp = 0.27 at 700 K
If Qp = Kp, the reaction is at equilibrium. Therefore, we can set up the equation and solve for the initial pressures of NO and Cl2.
Kp = 0.27 = (115)^2 / (PNO)^2 x (PCl2)
Since the initial pressures of NO and Cl2 are equal:
PNO = PCl2 = x (let's assume)
0.27 = (115)^2 / (x)^2 x (x)
Simplifying the equation:
0.27 = 13225 / x^4
Cross multiplying:
0.27x^4 = 13225
Dividing both sides by 0.27:
x^4 = 49018.52
Taking the fourth root of both sides:
x ≈ 16.91
Therefore, the initial pressures of NO and Cl2 are approximately 16.91 torr.
To find the initial pressures of NO and Cl2, we can use the equilibrium constant (Kp) along with the given partial pressure of NOCl at equilibrium.
Let's denote the initial pressure of NO and Cl2 as P(NO) and P(Cl2) respectively.
The balanced equation tells us that the stoichiometric ratio between NO and NOCl is 2:2. This means that for every 2 moles of NO that react, 2 moles of NOCl are produced.
Using the ideal gas law equation (PV = nRT), we can equate the initial and equilibrium partial pressure of NOCl and solve for the moles of NOCl.
Initial partial pressure of NOCl = P(NOCl) = 115 torr
Equilibrium partial pressure of NOCl = P(NOCl) = 115 torr
Since the stoichiometric ratio between NO and NOCl is 2:2, the moles of NOCl produced is equal to the moles of NO that reacted.
Therefore, n(NOCl) = 2 * n(NO)
Since the moles of a gas can be expressed using the ideal gas equation as n = PV / RT, we can substitute the equations for the moles of NOCl and NO.
2 * P(NOCl) = 2 * P(NO) * V / RT
Since the volume (V), temperature (T), and the gas constant (R) are constant, we can cancel them out from both sides of the equation.
2 * P(NOCl) = 2 * P(NO)
Now we can use the equilibrium constant (Kp) to relate the partial pressures of NOCl, NO, and Cl2.
Kp = P(NOCl)^2 / (P(NO) * P(Cl2))
Substituting the given values and the relationship between the partial pressures, we have:
0.27 = (115^2) / (P(NO) * P(Cl2))
Now we need to solve this equation for P(NO) and P(Cl2).
First, let's rearrange the equation:
P(NO) * P(Cl2) = (115^2) / 0.27
P(NO) * P(Cl2) = 554074.07
Since the initial partial pressures of NO and Cl2 are equal, we can substitute them with P(initial).
P(initial)^2 = 554074.07
Now we solve for P(initial):
P(initial) = √(554074.07)
P(initial) ≈ 744.26 torr
Therefore, the initial pressures of NO and Cl2 are approximately 744.26 torr.