2NO2(g) <-----> N2O4(g)
If 0.625mol of dinitrogen tetraoxide in 5.0L is allowed to come to equilibrium and at equilibrium has a concentration of 0.0750M, Calculate K for this reaction.
Does K= 7.5?
yes.
To calculate the equilibrium constant (K) for the given reaction, we need to use the given concentration of the reactant and the stoichiometry of the reaction. The balanced equation tells us that the molar ratio between NO2 and N2O4 is 2:1.
Given information:
Initial concentration of NO2 (dinitrogen tetraoxide) = 0.0750 M
Volume (V) = 5.0 L
Molar ratio (NO2 to N2O4) = 2:1
The equilibrium concentrations can be determined using the formula:
K = [N2O4] / ([NO2]^2)
To find the equilibrium concentration of N2O4, we can assume it to be "x" as it's not explicitly given. Since the molar ratio between NO2 and N2O4 is 2:1, the equilibrium concentration of NO2 will be 2x.
Re-writing the given equation in terms of concentrations:
2NO2(g) ⇌ N2O4(g)
Using the given information, we have:
[NO2] = (2 * x) = 2x
[N2O4] = x
At equilibrium, [NO2] = 0.0750 M, so we can substitute this value into the equation:
0.0750 = 2x
Solving this equation, we find:
x = 0.0375 M
With the calculated equilibrium concentrations, we can substitute them into the equilibrium constant expression:
K = [N2O4] / ([NO2]^2)
K = (0.0375) / ((0.0750)^2)
K ≈ 0.67
Therefore, the value of K for this reaction is approximately 0.67, not 7.5.