a 0.76 mol sample of pcl5 is placed in a 500 ml reaction vessel. what is the concentration of pcl5 when the reaction
pcl5(g) ⇀↽ pcl3(g) + cl2(g)
has reached equilibrium at 250◦c (when kc = 1.8)
kc= [x)(x)/(2*.76-x)= 1.8
x^2=1.8(1.52-x)
solve that quadratic for x
x^2+1.8x-2.74=0
x= (-1.8+-sqrt(3.24+10.96) )/2
x= -.9+- 3.77/2= ..985 moles/liter
or conce of Pcl5 is 2*.76-.985
check my math.
To find the concentration of PCl₅ at equilibrium, we'll need to use the given equilibrium constant (Kc) and the initial moles of PCl₅.
First, let's calculate the initial concentration of PCl₅:
We are given that the sample of PCl₅ has 0.76 moles. The volume of the reaction vessel is 500 mL, which is equal to 0.5 L. Therefore, the initial concentration of PCl₅ is:
[PCl₅] = moles/volume = 0.76 mol / 0.5 L = 1.52 mol/L
Now, we need to determine how the concentrations of PCl₃ and Cl₂ change at equilibrium. According to the balanced equation, 1 mole of PCl₅ produces 1 mole of PCl₃ and 1 mole of Cl₂. So, at equilibrium, the concentrations of PCl₃ and Cl₂ will also be equal.
Let's say the equilibrium concentration of PCl₃ and Cl₂ is x mol/L.
Using the equilibrium constant expression and the stoichiometry of the reaction, we can write the following equation:
Kc = [PCl₃] * [Cl₂] / [PCl₅]
Since [PCl₃] = x and [Cl₂] = x, we can rearrange the equation as:
Kc = x * x / 1.52
Now, substitute the given value of the equilibrium constant (Kc = 1.8) into the equation and solve for x:
1.8 = x² / 1.52
Multiply both sides of the equation by 1.52:
x² = 1.8 * 1.52
x² = 2.736
Taking the square root of both sides:
x = √2.736
x ≈ 1.656
Therefore, the equilibrium concentration of PCl₃, Cl₂, and PCl₅ is approximately 1.656 mol/L.