Answer to Question #271393 in General Chemistry for jena

Question

1. Consider the reaction: 2BrF_5 (g) <---> Br_{2 (g)} + 5F_{2 (g)}

• If the equilibrium concentrations of BrF₅, Br₂ and F₂ are 0.46 M, 1.2 M and 0.93 M, what is the value of K_c?

Answer 1

Balanced chemical equation:

2BrF5(g) ⇌ Br2(g) + 5F2(g)

The equilibrium expression (Kc) for the reaction:

Hence

We know that Kp = Kc (RT)Δn

Kc = 394.53

Convert 250∘C to Kelvin (K): T = 250∘C + 273.15 = 523.15 K

Δn = (total number of moles of products) - (total number of moles in reactants)

Δn = (5 + 1) - (2) = 6 - 2 = 4

Δn = 4

R = 0.08206 L atm mol^-1 K ^-1

Now, plug in all the numbers we found: Kp = Kc (RT)

Kp = 394.53 × (0.08206 × 523.15 K)4 = 1.34×10

Kp = 1.34×10