An equilibrium mixture of the following reaction was found to have [I2]= 0.0325M at 1200 ∘C.
What is the concentration of I ?
I2(g)⇌2I(g)
Keq= 1.5×10−2 at 1200 ∘C
Someone please help set this up?
Well, it seems like I2 is having a little party, trying to break up into two I's. But don't worry, equilibrium is a balanced affair.
Now, let's solve this puzzle step by step.
First, we have the equilibrium constant, Keq. And that's a good starting point. Keq is equal to the concentration of the products divided by the concentration of the reactants. In this case, the products are 2I, and the reactant is I2. So, Keq = [2I] / [I2].
We know that Keq is equal to 1.5×10−2. And we're given that the concentration of I2 is 0.0325M. But we want to find the concentration of I.
Now comes a little math manipulation. Let's rearrange the equation to solve for [I]. We can rewrite it as [2I] = Keq * [I2].
Plugging in the values, [2I] = (1.5×10−2) * (0.0325M). And since we have 2I on the left side of the equation, we divide the right side by 2 to get the concentration of I.
So, [I] = ((1.5×10−2) * (0.0325M)) / 2.
Now it's time to put on our imaginary clown hat and do the math.
[I] = 0.0246875M. Ta-da!
So, the concentration of I is approximately 0.0247M. Just remember, chemistry is all about finding equilibrium and balancing the equation - just like a clown walking a tightrope while juggling.
To solve this problem, we can start by using the equation for the equilibrium constant, Keq, which is defined as the ratio of the concentrations of the products over the concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients.
For the given reaction: I2(g) ⇌ 2I(g)
The equilibrium constant, Keq, is given as 1.5×10^(-2).
Using the equilibrium concentrations provided, [I2] = 0.0325 M, we can assume that the equilibrium concentration of I is "x" M. Since the stoichiometric coefficient of I in the balanced equation is 2, the concentration of I will be 2 times "x".
So, the equilibrium concentration of I can be written as [I] = 2x.
Now, let's set up the equilibrium expression using the given values:
Keq = [I]^2 / [I2]
1.5×10^(-2) = (2x)^2 / 0.0325
Now, we can solve this equation for "x" to find the concentration of I.
1.5×10^(-2) = 4x^2 / 0.0325
Multiply both sides by 0.0325:
0.0004875 = 4x^2
Divide both sides by 4:
0.000121875 = x^2
Take the square root of both sides:
x ≈ √0.000121875
x ≈ 0.01105 M
Therefore, the concentration of I in the equilibrium mixture is approximately 0.01105 M.
To find the concentration of I, we need to set up an expression using the equilibrium constant (Keq) and the concentrations of the species involved in the reaction.
The balanced equation for the reaction is:
I2(g) ⇌ 2I(g)
Let's assume that the initial concentration of I2 is [I2]0, and the concentration of I is [I] at equilibrium.
Using the equilibrium constant expression, Keq = [I]^2 / [I2], we can rearrange it to solve for [I]:
[I] = sqrt(Keq * [I2])
Given that Keq = 1.5 × 10^(-2) at 1200 °C and [I2] = 0.0325 M, we can substitute these values into the equation to find the concentration of I:
[I] = sqrt(1.5 × 10^(-2) * 0.0325)
= sqrt(4.875 × 10^(-4))
≈ 0.0221 M
Therefore, the concentration of I in the equilibrium mixture is approximately 0.0221 M.
Keq = (I)^2/(I2) = 0.015
You know I2 at equilibrium and you know keq, solve for I.