10 mole of nitrous oxide in 2dm3 container decompose to nitrogen and oxygen at equilibrium the concentration of nitrous oxide is 2.20 mole calculate kc
Well, if 10 moles of nitrous oxide decompose to nitrogen and oxygen, it means that at equilibrium, 10 moles of nitrous oxide will be converted into 10 moles of nitrogen and oxygen combined.
So, the equilibrium concentration of nitrous oxide would be 2.20 moles divided by the volume of the container, which is 2 dm3. This gives us a concentration of 1.10 moles/dm3.
Since the concentration of nitrogen and oxygen is equal to the concentration of nitrous oxide at equilibrium, the combined concentration would also be 1.10 moles/dm3.
Therefore, the equilibrium constant, Kc, would be [N2][O2] / [N2O], which is (1.10)(1.10) / 1.10.
And guess what? The answer is 1.10!
Isn't it amazing when an equilibrium constant turns out to be the same as the concentration of the reactants? It's almost like a magic trick!
To calculate the equilibrium constant (Kc) for the reaction, we need to use the balanced chemical equation for the decomposition of nitrous oxide (N2O):
N2O(g) ⇌ N2(g) + O2(g)
The given information tells us that the initial concentration of N2O is 2.20 moles in a 2 dm^3 container.
At equilibrium, we can assume that x moles of N2O decomposed, which leads to the formation of x moles of N2 and x moles of O2.
So, the equilibrium concentration of N2O is (2.20 - x) moles in a 2 dm^3 container.
The equilibrium concentrations of N2 and O2 are each x moles in a 2 dm^3 container.
Now, we can express the equilibrium constant (Kc) using the concentrations:
Kc = [N2] * [O2] / [N2O]
Substituting the concentrations into the equation:
Kc = (x * x) / (2.20 - x)
To solve for x, we need to set up an equation based on the given information:
2.20 - x = 10 - x
Simplifying the equation:
2.20 = 10
This means that x = 7.80 moles.
Now we can substitute this value into the equilibrium constant equation to find Kc:
Kc = (7.80 * 7.80) / (2.20 - 7.80)
Kc = (60.84) / (-5.60)
Kc ≈ -10.90 (approximately)
So, the equilibrium constant (Kc) for the given reaction is approximately -10.90.
To calculate the equilibrium constant (Kc) for the decomposition of nitrous oxide (N2O) into nitrogen (N2) and oxygen (O2), we need to know the balanced equation for the reaction.
The balanced equation for the decomposition of nitrous oxide is:
2 N2O(g) ⟶ 2 N2(g) + O2(g)
From the equation, we can see that the stoichiometric ratio between nitrous oxide and nitrogen is 2:2, and between nitrous oxide and oxygen is 2:1.
Given data:
- Initial concentration of nitrous oxide (N2O): 2.20 mole
- Volume of the container: 2 dm^3
To find the equilibrium concentration of nitrous oxide (N2O) after decomposition, we need to use the given initial concentration and the stoichiometric ratio. The equilibrium concentration of nitrous oxide (N2O) can be calculated as follows:
Initial concentration of N2O = 2.20 mole
Volume of the container = 2 dm^3
Concentration of N2O = (Initial moles of N2O) / (Volume of the container)
= 2.20 mole / 2 dm^3
= 1.10 mole/dm^3
Now, we have the equilibrium concentration of nitrous oxide (N2O) as 1.10 mole/dm^3.
To calculate the equilibrium constant (Kc), we need to use the equilibrium concentrations of all the species involved in the reaction. In this case, we have the concentration of nitrous oxide (N2O) as 1.10 mole/dm^3.
Kc = [(Concentration of N2)^2 * (Concentration of O2)] / (Concentration of N2O)^2
Since the stoichiometric ratio indicates that 2 moles of N2 are formed for every 2 moles of N2O that decompose, and 1 mole of O2 is formed for every 2 moles of N2O that decompose, we can determine the equilibrium concentrations of N2 and O2:
Concentration of N2 = 2 * Concentration of N2O
= 2 * 1.10 mole/dm^3
= 2.20 mole/dm^3
Concentration of O2 = 1 * Concentration of N2O
= 1 * 1.10 mole/dm^3
= 1.10 mole/dm^3
Now, substituting the values into the formula for Kc:
Kc = [(2.20 mole/dm^3)^2 * (1.10 mole/dm^3)] / (1.10 mole/dm^3)^2
Kc = (4.84 mole^2/dm^6 * mole) / (1.21 mole^2/dm^6)
Kc = 4
Therefore, the equilibrium constant (Kc) for the decomposition of nitrous oxide is 4.
(N2O) = 10 mols/2 dm3 = 5 Molar
..................2N2O ==> 2N2 + O2
I...................5................0........0
C..................-2x.............2x......x
E.................5-2x............2x.......x
Kc = (N2)^2(O2)/(N2O)
We have a problem here because of your post. Is that 2.20 mole in the problem 2.20 molar or is it 2.20 mole. I will assume it is 2.20 mol; however, you say that is a concentration and it can't be. If it is concentration it means 2.20 MOLAR. Again, I'll assume 2.20 mol N2O at equilibrium in the 2 L container so (N2O) at equilibrium will be 2.20/2 = 1.1 M.
So you know 5-2x = 1.1. Solve for x = 1.95 M. That means N2O is 1.1 M and O2 = x = 1.95 M and N2 is 2*1.95 = 3.9 M. Plug those values into the Kc expression bove and solve for Kc.
IF YOU MEAN that N2O at equilibrium is 2.20 molar, then
5-2x = 2.20, then x = 1.4. So in assumption, (N2O) = 2.20, (N2) = 2.8 and (O2) = 1.4. Plug those numbers into Kc expression and solve for Kc. Post your work if you get stuck. I hope this isn't confusing but you must learn to post problems correctly for it makes a difference it you the concentration is 2.20 M or if you have 2.20 mols.