I₂(g) ⇌ 2 I(g) The Kc of this reaction is 3.820 × 10⁻⁵ at 741.0 °C.

Question

Consider the reaction below: I₂(g) ⇌ 2 I(g) The Kc of this reaction is 3.820 × 10⁻⁵ at 741.0 °C. What is Kp at the same temperature for the reverse reaction? Note Kc is often called K. (R = 0.08206 L･atm/mol･K.) 2 I(g) ⇌ I₂(g).

I've tried using the formula with no luck, I'm not sure what I'm getting incorrect! Thanks so much or any help!

Answer 1

Hi Terra H.!

We are given the information for the following chemical equation:

I₂(g) <------> 2I(g) Kc = 3.820x10^-5

For the forward direction, the Kc takes the following form:

Kc = [I]²/[I₂] = 3.820x10^-5

However, we are interested in the reverse reaction. In that case, the equilibrium constant takes the following form:

Kc^' = [I₂]/[I]² = 1/Kc = 1/3.820x10^-5

Kc^' = 2.618x10⁴

Kp = Kc^'(RT)^Δn

=> Δn = n_products - n_reactants <=> 1 - 2 => Δn = = -1

NOTE: Δn represents the change in the number of moles between products and reactants. In this case, we are going backwards, so I₂ is our product and I is our reactants.

<=> (2.618x10⁴)(0.08206 L*atm*mol^-1*K^-1 * 1014K)^-1

Kp = 314.6

Let me know if that works out any better for you!

Cheers