The equilibrium constant is 16 for the gas phase reaction SO2 + NO2 <--> SO3 + NO at a certain temperature. If 3.0 moles each of SO2 and NO2 are placed together in an empty 1.0 liter flask and the system is allowed to come to equilibrium at this temperature, what will be the concentration of SO3 at equilibrium?
(I came up with 2.4 M, but I'm not sure if I set the problem up correctly).
I obtained 2.4 also.
can someone help me out?
at 25¡ãc,the value of k�0Š4 is 1.7x10^8.which speciesis present in greater amount at equilibrium -so3(g)or so2(g).why?
Science
To find the concentration of SO3 at equilibrium, we first need to set up the reaction quotient, Q, which is the ratio of the concentrations of the products to the concentrations of the reactants at any given point during the reaction.
The balanced equation for the reaction is:
SO2 + NO2 ⇌ SO3 + NO
According to the given information, the equilibrium constant, K, is 16. The equilibrium expression for this reaction is:
K = [SO3][NO] / [SO2][NO2]
Since the reaction has not yet reached equilibrium, we will use the stoichiometry of the reaction to determine the initial concentrations of the reactants:
Initial concentration of SO2 = 3.0 moles / 1.0 L = 3.0 M
Initial concentration of NO2 = 3.0 moles / 1.0 L = 3.0 M
At equilibrium, the concentrations of the reactants (SO2 and NO2) will decrease, while the concentrations of the products (SO3 and NO) will increase. Let's assume that x moles of SO3 are formed.
Therefore, at equilibrium:
[SO2] = 3.0 M - x
[NO2] = 3.0 M - x
[SO3] = x
[NO] = x
Now, we can plug these values into the equilibrium expression:
K = [SO3][NO] / [SO2][NO2]
16 = x*x / (3.0 - x)(3.0 - x)
Simplifying the equation gives:
16(9 - x^2) = x^2
144 - 16x^2 = x^2
17x^2 = 144
x^2 = 8.471
Taking the square root of both sides, we find that x ≈ 2.91
Therefore, the concentration of SO3 at equilibrium is approximately 2.91 M.