Given: 2N2(g) +O(2) --> 2N2O(g) Kc =1.2x10^(-35)
N2O4(g) --> 2NO2(g) Kc = 4.6x10^(-3)
1/2N2(g) + O2(g) --> NO2(g) Kc= 4.1x10^(-9)
Calculate Kc for the reaction between one mole of dinitrogen oxide gas and oxygen gas to give nitrogen tetroxide gas.
I got 1054.9 as my answer. I reversed the first eq and did (1/K1^(1/2)) added it to reverse of second eq (1/K2) then added that to the 3rd eq times 2 (K3^2). This got me the right equation. Do I just multiply the new Kc values together to get the final Kc of the new equation?
To calculate the equilibrium constant (Kc) for the reaction between one mole of dinitrogen oxide gas (N2O) and oxygen gas (O2) to give nitrogen tetroxide gas (N2O4), we can use the equilibrium constant expression and manipulate the given equilibrium constants.
The balanced chemical equation for the reaction is:
1/2N2O(g) + O2(g) ⟶ N2O4(g)
We can write the equilibrium constant expression as:
Kc = [N2O4] / ([1/2N2O] * [O2])
From the given equilibrium constants, we can manipulate them to match the desired reaction equation.
Given:
2N2(g) + O2(g) ⟶ 2N2O(g) Kc = 1.2x10^(-35)
N2O4(g) ⟶ 2NO2(g) Kc = 4.6x10^(-3)
1/2N2(g) + O2(g) ⟶ NO2(g) Kc = 4.1x10^(-9)
1) Multiply the second equation by 1/2:
1/2(N2O4(g) ⟶ 2NO2(g)) ⟹ 1/4(N2O4(g)) ⟶ (NO2(g))^2
2) Add the modified equation (2) and equation (3):
1/4(N2O4(g)) + (NO2(g))^2 + 1/2N2(g) + O2(g) ⟶ (NO2(g))^2 + NO2(g)
1/4(N2O4(g)) + 1/2N2(g) + O2(g) ⟶ (NO2(g))^2 + NO2(g)
3) Multiply the previous equation by 2:
2(1/4(N2O4(g)) + 1/2N2(g) + O2(g)) ⟶ 2(NO2(g))^2 + 2NO2(g)
Now we have:
1/2N2O(g) + O2(g) ⟶ 2(N2O4(g)) + 2NO2(g)
Since the coefficients of N2O and O2 in the modified equation match the desired reaction equation (1/2N2O + O2 ⟶ N2O4), we can substitute the equilibrium constants accordingly.
Kc(desired) = (Kc(1))^(-1) * (Kc(2))^2 * (Kc(3))^2
Substituting the given equilibrium constants:
Kc = (1.2x10^(-35))^(-1) * (4.6x10^(-9))^2 * (4.1x10^(-9))^2
Calculating this expression gives Kc = 4.118 x 10^(-63).
To calculate the equilibrium constant (Kc) for the reaction between one mole of dinitrogen oxide gas (NO) and oxygen gas (O2) to give nitrogen tetroxide gas (N2O4), you need to utilize the given equilibrium constants for the reactions involving the same species.
First, let's set up the balanced equation:
1/2N2(g) + O2(g) --> NO2(g)
2NO2(g) --> N2O4(g)
Now, we can combine these two equations to get the desired reaction:
1/2N2(g) + O2(g) --> NO2(g) + NO2(g)
1/2N2(g) + O2(g) --> N2O4(g)
To find Kc for the desired reaction, we multiply the Kc values for the individual reactions since the desired reaction is the combination of the two:
Kc(desired) = Kc(1/2N2 + O2 --> NO2) * Kc(2NO2 --> N2O4)
Now, substitute the given equilibrium constants into the equation:
Kc(desired) = 4.1x10^(-9) * 4.6x10^(-3)
Multiply the values:
Kc(desired) = 1.886x10^(-11)
Therefore, the equilibrium constant (Kc) for the reaction between one mole of dinitrogen oxide gas and oxygen gas to give nitrogen tetroxide gas is approximately 1.886x10^(-11).