Ask Question
23 May, 01:40

For the reaction i2 (g) + br2 (g) ←-→2ibr (g), kc=280 at 150 ∘c. suppose that 0.520 mol ibr in a 2.00-l flask is allowed to reach equilibrium at 150 ∘c. part a what is the equilibrium concentration of ibr?

+2
Answers (1)
  1. 23 May, 05:10
    0
    First, we have to get the initial [IBr] = 0.520 mol / 2 L = 0.26 M

    According to the reaction balanced equation:

    and using ICE table:

    I2 (g) + Br2 (g) ↔ 2IBr (g)

    initial 0 0 0.26

    Change + X + X - 2X

    Equ X X (0.26-2X)

    when Kc = [IBr]^2/[I2][Br2]

    so by substitution:

    280 = (0.26-2X) ^2 / X^2 by solving this equation for X

    ∴X = 0.0139

    ∴[I2] = [Br]2 = 0.0139

    and [IBr] = 0.26 - (2*0.0139)

    = 0.2322 M
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “For the reaction i2 (g) + br2 (g) ←-→2ibr (g), kc=280 at 150 ∘c. suppose that 0.520 mol ibr in a 2.00-l flask is allowed to reach ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers