A sealed 1.0L flask is charged with .500mol I2 and .500mol Br2 and an equilibrium reaction esues: I2(g)+Br2(g)=2IBr(g)
When the container contents achieve equilibrium, the flask contains .84 mol of IBr. What is the value of the eqilibrium constant? (K)
I2(g)+Br2(g)=2IBr(g)
I .500 .500 0
C -x -x +2x
E .500-x .500-x .84
I know K=[product]/[reactant] but I don't know what to do with the information from there.
Never mind I figured it out
ths results cm 4rm table..products/reactants..(84)^2/(416)^2(416)^2
To find the value of the equilibrium constant (K), we need to use the concentrations of the reaction species at equilibrium. Given that the flask contains 0.84 mol of IBr, we can calculate the concentrations of I2 and Br2 at equilibrium.
The balanced equation for the reaction is:
I2(g) + Br2(g) ⇌ 2IBr(g)
Let's denote the initial concentrations of I2 and Br2 as [I2]₀ and [Br2]₀, respectively. As the initial amounts of I2 and Br2 were both 0.500 mol, their initial concentrations are also 0.500 M.
At equilibrium, the number of moles of IBr is 0.84, which corresponds to a concentration of 0.84 M.
Since 2 moles of IBr are formed for each mole of I2 or Br2 consumed, the change in molar concentration for I2 and Br2 is -0.42 M (0.84 mol * 2) because two moles are consumed.
Therefore, the equilibrium concentrations become:
[I2] = [I2]₀ - 0.42 M
[Br2] = [Br2]₀ - 0.42 M
Now that we have the equilibrium concentrations of I2, Br2, and IBr, we can substitute these values into the equilibrium constant expression:
K = ([IBr] / [I2]) * ([IBr] / [Br2])
Using the given values of [IBr] = 0.84 M, [I2]₀ = [Br2]₀ = 0.500 M, and [IBr] = [I2]₀ - 0.42 M = [Br2]₀ - 0.42 M, we can plug these values into the equation:
K = (0.84 M / (0.500 M - 0.42 M)) * (0.84 M / (0.500 M - 0.42 M))
Simplifying further:
K = (0.84 M / 0.080 M) * (0.84 M / 0.080 M) = 110.25
Therefore, the value of the equilibrium constant (K) is 110.25.