At a particular temperature, K = 4.0 x 10-7 for thereaction:N2O4(g) <-->2NO2(g)In an experiment, 1.0 mol N2O4 is placedin a 10 L vessel. Calculate the concentrations ofN2O4 and NO2 when this reactionreaches equilibrium.
Question
At a particular temperature, K = 4.0 x 10-7 for thereaction:N2O4(g) <-->2NO2(g)In an experiment, 1.0 mol N2O4 is placedin a 10 L vessel. Calculate the concentrations ofN2O4 and NO2 when this reactionreaches equilibrium.
Solution
Sure, let's solve this step by step.
Step 1: Write the equilibrium expression for the reaction.
The equilibrium constant expression for the reaction N2O4(g) <--> 2NO2(g) is K = [NO2]^2 / [N2O4].
Step 2: Write the initial and change in concentrations.
Initially, we have 1.0 mol of N2O4 in a 10 L vessel, so the initial concentration of N2O4 is 1.0 mol / 10 L = 0.1 M. The initial concentration of NO2 is 0.
As the reaction proceeds to equilibrium, the concentration of N2O4 decreases by x and the concentration of NO2 increases by 2x (because there are 2 moles of NO2 for every mole of N2O4 that reacts).
So at equilibrium, [N2O4] = 0.1 - x and [NO2] = 2x.
Step 3: Substitute the equilibrium concentrations into the equilibrium expression and solve for x.
4.0 x 10^-7 = (2x)^2 / (0.1 - x)
Solving this equation for x gives x = 0.002 M.
Step 4: Substitute x back into the equilibrium concentrations to find the concentrations of N2O4 and NO2 at equilibrium.
[N2O4] = 0.1 - 0.002 = 0.098 M [NO2] = 2 * 0.002 = 0.004 M
So, the equilibrium concentrations of N2O4 and NO2 are 0.098 M and 0.004 M, respectively.
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