Question

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol of Cl2 in a 1.00-L reaction vessel at 230°C, what is the concentration of PCl3 when equilibrium has been established?

Answer 1

The concentration of PCl3 at equilibrium is 24.01 M.

Step-by-step explanation:

To calculate the concentration of PCl3 at equilibrium, we can use the equilibrium constant expression:

Kc = [PCl5] / ([PCl3] * [Cl2])

Given that Kc = 49 and the initial concentrations of PCl3 and Cl2 are both 0.70 M, we can set up the equation as:

49 = [PCl5] / (0.70 * 0.70)

Solving for [PCl5], we find:

[PCl5] = 0.70 * 0.70 * 49

[PCl5] = 24.01 M

Therefore, the concentration of PCl3 at equilibrium is 24.01 M.

Answer 2

Concentration of PCl₅ at equilibrium = 24.01 mole/lit

Step-by-step explanation:

Given

Equilibrium constant Kc = 49

PCl₃ (g) + Cl₂(g) ⇄ PCl₅(g)

According to Law of Mass action

Given that

Concentration of PCl₃ = 0.7 mole/lit & conc. of Cl₂ = 0.7 mole/lit

∴ volume of the vessel constant = 1 lit

`K_(c)=([PCl_(5) ])/([PCl_(3) ][Cl_(2) ]) \\\\49= (x)/(0.7X0.7)`

∴ Concentration of PCl₅ at equilibrium = 49 x 0.7 x 0.7 = 24.01