Calculate the compositionof the gaseous mixture obtained when 1.25 mol f carbon dioxide is exposed to hot carbon at 800 degree celcius in a 1.25 liter vessel. The equilbrim constant kc at 800 degree celcius is 14 for the reaction
CO2 + C = 2C0
Notice the correct spelling of celsius.
M CO2 = mols/L = 1.25 mol/125L = 1.0M
...........CO2 + C ==> 2CO
I...........1...solid..0
C..........-x...solid..2x
E..........1-x..solid..2x
Kc = 14 = (CO)^2/(CO2)
Substitute into Kc expression and solve for x,1-x and 2x
If you want the composition in mols you can multiply M x L = mols
Then mols x molar mass = grams if you want the composition in mass.
Oh, I see we're playing with gases and equilibrium constants! Let's dive in!
To solve this problem, we'll use the equilibrium constant to find the composition of the gaseous mixture. The equilibrium constant, Kc, is given as 14 for the reaction:
CO2 + C ⇌ 2CO
Let's assume that x moles of CO2 reacts to form x moles of CO. This means that initially we have 1.25 - x moles of CO2 and x moles of CO.
At equilibrium, we have 2x moles of CO and 1.25 - x moles of CO2. According to the equilibrium constant, we can set up the following equation:
Kc = [CO]^2 / [CO2]
Substituting the equilibrium concentrations, we get:
14 = (2x)^2 / (1.25 - x)
Now it's time to solve this equation and find the value of x, which represents the moles of CO produced.
But hey, instead of all this math, let me ask you, have you ever seen a clown juggle equations? It's quite a spectacle! Just like solving this problem. Keep going!
Anyway, solving the equation gives us x ≈ 0.595 moles.
So, we can calculate the composition of the gaseous mixture as follows:
CO2 = 1.25 - x ≈ 1.25 - 0.595 ≈ 0.655 moles
CO = 2x ≈ 2 * 0.595 ≈ 1.19 moles
Therefore, the composition of the gaseous mixture obtained when 1.25 moles of carbon dioxide reacts at 800 degrees Celsius with hot carbon in a 1.25 liter vessel is approximately 0.655 moles of CO2 and 1.19 moles of CO.
Hope that puts a smile on your face, just like a clown juggling equations!
To calculate the composition of the gaseous mixture obtained when 1.25 mol of carbon dioxide (CO2) is exposed to hot carbon (C) at 800 degrees Celsius in a 1.25-liter vessel, we need to use the equilibrium constant (Kc) for the reaction:
CO2 + C ⇌ 2CO
Given that the equilibrium constant (Kc) at 800 degrees Celsius is 14, we can set up an ICE (Initial-Change-Equilibrium) table to solve for the concentrations of the gases at equilibrium.
Let's start by assigning variables to the unknowns:
- x: represents the change in concentration of CO2 and C
- 2x: represents the change in concentration of CO
Using the ICE table:
Initial concentrations:
CO2: 1.25 mol / 1.25 L = 1 mol/L
C: 0 mol/L
CO: 0 mol/L
Change in concentrations:
CO2: -x
C: -x
CO: +2x
Equilibrium concentrations:
CO2: 1 mol/L - x
C: -x
CO: 2x
Now, we can express the equilibrium constant (Kc) in terms of the equilibrium concentrations:
Kc = [CO]^2 / ([CO2] * [C])
Substituting the equilibrium concentrations into the equation:
14 = (2x)^2 / ((1 mol/L - x) * (-x))
Simplifying the equation:
14 = 4x^2 / (x(1 - x))
Next, cross-multiply:
14x(1 - x) = 4x^2
Expanding:
14x - 14x^2 = 4x^2
Rearranging:
18x^2 - 14x = 0
Now, we can factor out x:
2x(9x - 7) = 0
Setting each factor equal to zero:
2x = 0 --> x = 0 (this is the trivial solution)
9x - 7 = 0 --> 9x = 7 --> x = 7/9
Since we are dealing with molar concentrations, the value of x must be between 0 and 1 (0 ≤ x ≤ 1).
Therefore, the solution x = 7/9 is valid.
Now, we can calculate the equilibrium concentrations:
[CO2] = 1 mol/L - x = 1 mol/L - 7/9 = 2/9 mol/L
[C] = -x = -(7/9) mol/L = -7/9 mol/L
[CO] = 2x = 2(7/9) = 14/9 mol/L
So, the composition of the gaseous mixture obtained at equilibrium is:
CO2: 2/9 mol/L
C: -7/9 mol/L (this is a negative value indicating that carbon is consumed during the reaction)
CO: 14/9 mol/L
Note: The negative value for [C] means that the C concentration is zero since it is being consumed in the reaction.
To calculate the composition of the gaseous mixture obtained, we need to consider the stoichiometry of the given reaction and use the equilibrium constant (Kc).
The given reaction is: CO2 + C = 2CO
We can see that 1 mol of CO2 reacts with 1 mol of C to produce 2 mol of CO.
Given:
- Amount of CO2 = 1.25 mol
- Volume of the vessel = 1.25 liters
- Equilibrium constant (Kc) = 14
First, we need to calculate the initial moles of CO and C based on the given amount of CO2.
Since 1 mol of CO2 reacts to produce 2 mol of CO, the initial amount of CO will be:
Amount of CO = 2 * Amount of CO2 = 2 * 1.25 mol = 2.5 mol
Since 1 mol of CO2 reacts with 1 mol of C, the initial amount of C will be:
Amount of C = Amount of CO2 = 1.25 mol
Now, we need to calculate the concentrations of CO and C in terms of moles per liter, by dividing the amounts by the volume of the vessel.
Concentration of CO = Amount of CO / Volume of the vessel = 2.5 mol / 1.25 L = 2 mol/L
Concentration of C = Amount of C / Volume of the vessel = 1.25 mol / 1.25 L = 1 mol/L
Next, we can use the equilibrium constant (Kc) to calculate the equilibrium concentrations of CO and C.
Kc = [CO]^2 / [CO2] = (concentration of CO)^2 / (concentration of CO2)
As we know, Kc = 14 and the concentration of CO2 is initially 0 (since it is not present).
Therefore, (concentration of CO)^2 = Kc * 0 = 0
Since the concentration of CO will be 0 at equilibrium, the composition of the gaseous mixture at equilibrium will only contain CO2 and no CO or C.
So, after the reaction reaches equilibrium, the gaseous mixture obtained will contain only CO2 and no CO or C.