for N2O4(g)-><-2NO2(g) the value of K at 25 degrees C is 7.19*10-3. calculate N2O4 at equilibrium when NO2 =2.80*10-2 mol/L. Calculate NO2 at equilibrium when N2O4=7.00*10-2 mol/L
To solve this problem, we can use the expression for the equilibrium constant (K) and the given concentrations to find the unknown concentration.
1. Calculate N2O4 at equilibrium when NO2 = 2.80 * 10^-2 mol/L:
The equilibrium constant expression for the given reaction is:
K = [NO2]^2 / [N2O4]
We are given:
K = 7.19 * 10^-3
[NO2] = 2.80 * 10^-2 mol/L
Let's assume that the equilibrium concentration of N2O4 is represented by x mol/L. Since the stoichiometric ratio between N2O4 and NO2 is 1:2, the concentration of NO2 at equilibrium will be 2x mol/L.
Substituting these values into the equilibrium constant expression:
7.19 * 10^-3 = (2x)^2 / x
Rearranging the equation:
7.19 * 10^-3 = 4x^2 / x
Simplifying the equation:
7.19 * 10^-3 = 4x
Dividing by 4:
x = (7.19 * 10^-3) / 4
x = 1.798 * 10^-3 mol/L
Therefore, the concentration of N2O4 at equilibrium is 1.798 * 10^-3 mol/L when [NO2] = 2.80 * 10^-2 mol/L.
2. Calculate NO2 at equilibrium when N2O4 = 7.00 * 10^-2 mol/L:
Using the same equilibrium constant expression:
K = [NO2]^2 / [N2O4]
We are given:
K = 7.19 * 10^-3
[N2O4] = 7.00 * 10^-2 mol/L
Let's assume that the equilibrium concentration of NO2 is represented by x mol/L. Since the stoichiometric ratio between N2O4 and NO2 is 1:2, the concentration of N2O4 at equilibrium will be 0.5x mol/L.
Substituting these values into the equilibrium constant expression:
7.19 * 10^-3 = x^2 / (0.5x)
Rearranging the equation:
7.19 * 10^-3 = 2x
Dividing by 2:
x = (7.19 * 10^-3) / 2
x = 3.595 * 10^-3 mol/L
Therefore, the concentration of NO2 at equilibrium is 3.595 * 10^-3 mol/L when [N2O4] = 7.00 * 10^-2 mol/L.
To solve this problem, we will use the equilibrium expression and the given value of K.
The balanced equation for the reaction is:
N2O4(g) ⇌ 2NO2(g)
The equilibrium constant expression is:
K = [NO2]^2 / [N2O4]
Let's calculate the concentration of N2O4 at equilibrium when [NO2] = 2.80 * 10^(-2) mol/L.
Given:
K = 7.19 * 10^(-3)
[NO2] = 2.80 * 10^(-2) mol/L
Using the equilibrium constant expression, we can rearrange it to solve for [N2O4]:
K = [NO2]^2 / [N2O4]
[N2O4] = [NO2]^2 / K
Plugging in the values:
[N2O4] = (2.80 * 10^(-2))^2 / 7.19 * 10^(-3)
Calculating:
[N2O4] = 1.097 * 10^(-1) mol/L
Therefore, the concentration of N2O4 at equilibrium is approximately 1.097 * 10^(-1) mol/L.
Now, let's calculate the concentration of NO2 at equilibrium when [N2O4] = 7.00 * 10^(-2) mol/L.
Given:
K = 7.19 * 10^(-3)
[N2O4] = 7.00 * 10^(-2) mol/L
Again, using the equilibrium constant expression, we can solve for [NO2]:
K = [NO2]^2 / [N2O4]
[NO2]^2 = K * [N2O4]
[NO2] = √(K * [N2O4])
Plugging in the values:
[NO2] = √(7.19 * 10^(-3) * 7.00 * 10^(-2))
Calculating:
[NO2] ≈ 2.32 * 10^(-2) mol/L
Therefore, the concentration of NO2 at equilibrium is approximately 2.32 * 10^(-2) mol/L.
K=[NO2]^2/[N2O4]
a) you know K, you know N2O4. Calculate.
b) again, you know two , calculate the unknown.