Answer:
[Cl₂] in equilibrium is 1.26 M
Explanation:
This is the equilibrium:
PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)
Kc = 91
So let's analyse, all the process:
PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)
Initially 0.24 M 1.50M 0.12 M
React x x x
Some amount of compound has reacted during the process.
In equilibrium we have
0.24 - x 1.50 - x 0.12 + x
As initially we have moles of product, in equilibrium we have to sum them.
Let's make the expression for Kc
Kc = [PCl₅] / [Cl₂] . [PCl₃]
91 = (0.12 + x) / (0.24 - x) ( 1.50 - x)
91 = (0.12 + x) / (0.36 - 0.24x - 1.5x + x²)
91 (0.36 - 0.24x - 1.5x + x²) = (0.12 + x)
32.76 - 158.34x + 91x² = 0.12 +x
32.64 - 159.34x + 91x² = 0
This a quadratic function:
a = 91; b= -159.34; c = 32.64
(-b +- √(b² - 4ac)) / 2a
Solution 1 = 1.5
Solution 2 = 0.23 (This is our value)
So [Cl₂] in equilibrium is 1.50 - 0.23 = 1.26 M
