The equilibrium constant for the conversion of cis-stilbene to trans-stilbene is Kc = 24.0 at 200 C. If cis-stilbene is placed in an empty container at an initial concentration of 0.850 mol/L. What is the final concentration of trans-stilbene in the container at equilbrium?
A diagram illustrates cis-stilbene on the left and trans stilbene on the right with a reverse arrow <===> between
The answer is 0.816 mol L-1. What is the formula used and maths steps please.
Thanks
Cis ==> Trans
initial:
Cis = 0.850
trans = 0
change:
trans: +x
cis: -x
equilibrium:
cis = 0.850
trans = +x
Kc = 24.0 = (trans)/(cis)
24.0 = (x)/(0.850-x)
solve for x.
Ah, chemistry! The land of compounds and equations. Now, let's have some fun with your question.
To find the final concentration of trans-stilbene at equilibrium, we'll use the equilibrium constant (Kc) and take advantage of the stoichiometry of the reaction.
The chemical equation for the conversion of cis-stilbene to trans-stilbene can be written as follows:
cis-stilbene <===> trans-stilbene
We'll assume that both compounds are present in the same container, which means they'll have the same initial concentration (0.850 mol/L). Let's say the final concentration of trans-stilbene is x mol/L. Since the stoichiometry of the reaction is 1:1, the final concentration of cis-stilbene will also be (0.850 - x) mol/L.
The equilibrium constant (Kc) is given as 24.0, which means:
Kc = [trans-stilbene] / [cis-stilbene]
Now, we can substitute the concentrations into the equation:
24.0 = x / (0.850 - x)
To solve for x, let's first simplify the equation:
24.0 * (0.850 - x) = x
Now, let's distribute the 24.0:
20.4 - 24x = x
Bring the x terms to one side:
25x = 20.4
Divide both sides by 25:
x = 0.816
So, the final concentration of trans-stilbene at equilibrium is 0.816 mol/L. Voila!
Remember, this is just a mathematical representation of the relationship between the concentration of cis-stilbene and trans-stilbene at equilibrium. If you actually see a clown hop out of a flask and start doing calculations, I'd be worried.
To solve this question, we'll use the equilibrium constant expression for the given reaction:
Kc = [trans-stilbene] / [cis-stilbene]
Given that Kc = 24.0, and the initial concentration of cis-stilbene is 0.850 mol/L, let's assume the final concentration of trans-stilbene is x mol/L.
Using the equilibrium constant expression, we have:
24.0 = x / 0.850
To find the value of x, we can cross-multiply and solve for x:
24.0 * 0.850 = x
x = 20.4 mol/L
Therefore, the final concentration of trans-stilbene at equilibrium is 20.4 mol/L, which is approximately 0.816 mol/L.
The formula used is the equilibrium constant expression (Kc) and the mathematical steps involve solving the equation for the unknown variable (x) using cross-multiplication.
To solve this problem, we will use the equilibrium constant expression and the information provided.
The equilibrium constant expression for the given reaction is:
Kc = [trans-stilbene] / [cis-stilbene]
Given:
Kc = 24.0
Initial concentration of cis-stilbene ([cis-stilbene]initial) = 0.850 mol/L
Let's assume the final concentration of trans-stilbene at equilibrium is [trans-stilbene]x.
Using the equilibrium constant expression, we can write:
24.0 = [trans-stilbene]x / 0.850
To find [trans-stilbene]x, we can rearrange the equation:
[trans-stilbene]x = 24.0 * 0.850
[trans-stilbene]x = 20.4 mol/L
Therefore, the final concentration of trans-stilbene at equilibrium is 0.816 mol/L.