Answer to Question #201142 in Chemistry for Mark

Question

1.  Consider the given reaction.

2BrF5(g)  ⇌ Br2(g) + 5F2(g)               

If the respective equilibrium concentrations of BrF5, Br2, and F2 at 250C are 0.046M, 1.2 M, and 0.93M, what are the values of Kc and Kp?

Answer 1

Solution:

Balanced chemical equation:

2BrF₅(g) ⇌ Br₂(g) + 5F₂(g)

The equilibrium expression (K_c) for the reaction:

Therefore,

We know that K_p = K_c (RT)^Δn

K_c = 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: K_p = K_c (RT)^Δn

K_p = 394.53 × (0.08206 × 523.15 K)⁴ = 1.34×10⁹

K_p = 1.34×10⁹

Answer:

K_c = 394.53

K_p = 1.34×10⁹