A researcher put 10.0 moles of N2O into a 2-L container at some temperature where it decomposes according to the following: N2O = 2N2+ O2. At equilibrium 2.20 moles of N2O remain. Calculate the Kc for the reaction.
Kc = 24.5
Why did the nitrogen molecule go to therapy?
Because it couldn't handle the pressure of being in a container with so much decomposition going on!
In order to calculate the Kc for the reaction, we need to get serious for a moment. The equation for Kc is:
Kc = [N2]^2 * [O2] / [N2O]^2
From the given information, we know that at equilibrium, 2.20 moles of N2O remain. The reaction also tells us that for every 1 mole of N2O that decomposes, 2 moles of N2 and 1 mole of O2 are formed. Therefore, we can determine that 10.0 - 2.20 = 7.80 moles of N2O has decomposed.
Since 2 moles of N2 are formed for every 1 mole of N2O, we can say that 7.80 moles of N2O leads to the formation of 2 * 7.80 = 15.60 moles of N2.
The reaction also tells us that 1 mole of O2 is formed for every 1 mole of N2O, so 7.80 moles of N2O leads to the formation of 7.80 moles of O2.
Now, let's substitute these values into the Kc equation:
Kc = (15.60)^2 * (7.80) / (2.20)^2
Calculating this will give you the value of Kc. Time to whip out your calculator and crunch those numbers!
To calculate the equilibrium constant, Kc, for the given reaction, we need to use the stoichiometric coefficients of the balanced equation.
The given reaction is: N2O = 2N2 + O2
At the start, we have 10.0 moles of N2O in a 2-L container. After reaching equilibrium, 2.20 moles of N2O remain.
To find the concentrations of N2O, N2, and O2 at equilibrium, we need to convert the moles of each component into molar concentrations.
1) Concentration of N2O (initial):
Molar concentration = moles/volume = 10.0 moles/2 L = 5.0 M
2) Concentration of N2O (at equilibrium):
Molar concentration = moles/volume = 2.20 moles/2 L = 1.10 M
3) Concentration of N2 (at equilibrium):
We have 2N2 formed for every 1 mole of N2O that decomposes.
Since 2.20 moles of N2O remained, we can use the stoichiometric ratio to find the moles of N2 formed.
Moles of N2 = 2 * 2.20 moles = 4.40 moles
Molar concentration = moles/volume = 4.40 moles/2 L = 2.20 M
4) Concentration of O2 (at equilibrium):
We have 1 mole of O2 formed for every 1 mole of N2O that decomposes.
Since 2.20 moles of N2O remained, we can use the stoichiometric ratio to find the moles of O2 formed.
Moles of O2 = 1 * 2.20 moles = 2.20 moles
Molar concentration = moles/volume = 2.20 moles/2 L = 1.10 M
Now, we can write the expression for the equilibrium constant (Kc) using the molar concentrations:
Kc = [N2]^2 * [O2] / [N2O]^2
= (2.20)^2 * (1.10) / (1.10)^2
= 2.42
Therefore, the value of Kc for the given reaction is approximately 2.42.
To calculate the equilibrium constant (Kc) for the reaction, we need to determine the concentrations of the reactants and products at equilibrium.
Given:
- Initial moles of N2O: 10.0 moles
- Moles of N2O at equilibrium: 2.20 moles
To find the moles of N2 and O2 at equilibrium, we use the stoichiometry of the reaction: N2O = 2N2 + O2.
Since 1 mole of N2O produces 2 moles of N2 and 1 mole of O2, we can calculate the moles of N2 and O2 at equilibrium as follows:
Moles of N2 at equilibrium:
= 2 × (Initial moles of N2O - Moles of N2O at equilibrium)
= 2 × (10.0 moles - 2.20 moles)
= 2 × 7.80 moles
= 15.60 moles
Moles of O2 at equilibrium:
= 1 × (Initial moles of N2O - Moles of N2O at equilibrium)
= 1 × (10.0 moles - 2.20 moles)
= 1 × 7.80 moles
= 7.80 moles
Now, we can calculate the concentrations (in moles per liter) of each species at equilibrium.
Concentration of N2O at equilibrium:
= Moles of N2O at equilibrium / Volume of container
= 2.20 moles / 2 L
= 1.10 M
Concentration of N2 at equilibrium:
= Moles of N2 at equilibrium / Volume of container
= 15.60 moles / 2 L
= 7.80 M
Concentration of O2 at equilibrium:
= Moles of O2 at equilibrium / Volume of container
= 7.80 moles / 2 L
= 3.90 M
Now we can plug these concentrations into the equilibrium expression for Kc:
Kc = [N2]^2 × [O2] / [N2O]
= (7.80 M)^2 × (3.90 M) / (1.10 M)
= 237.48 M^2
Therefore, the value of Kc for the given reaction is 237.48 M^2.
You didn't balance the equation.
Set up an ICE chart. (N2O) initially = 10.0moles/2L = 5M
............2N2O ==> 2N2 + O2
initial.....10.0M....0......0
change.......-x......+2x...+x
equilibrium..10.00-x..2x....x
Kc = (N2)^2(O2)/(N2O)^2
You are given that at equilibrium the N2O is 2.20 moles.
10.00-x = 2.20, solve for x and complete the table. Convert moles to M in all cases and substitute into Kc expression and solve for Kc.