Ask Question
9 February, 00:55

Consider the reaction below that has a Keq of 8.98 X 10-2. The initial concentration of A and B is 1.68 M. Calculate the equilibrium concentration of C.

A + B



2C

+2
Answers (2)
  1. 9 February, 02:53
    0
    The equilibrium concentration of C is 0.4378 M

    Explanation:

    Step 1: Data given

    Keq = 8.98 * 10^-2

    The initial concentration A = 1.68 M

    The initial concentration B = 1.68 M

    Step 2: The balanced equation

    A + B ↔ 2C

    Step 3: The initial concentration

    [A] = 1.68 M

    [B] = 1.68 M

    [C] = O M

    Step 4: The concentration at the equilibrium

    [A] = 1.68 - X

    [B] = 1.68 - X

    [C] = 2X

    Step 5: Define Kc

    Kc = [C]² / [A][B]

    0.0898 = (2X) ² / (1.68 - X) (1.68 - X)

    X = 0.2189

    [A] = 1.68 - 0.2189 = 1.4611 M

    [B] = 1.68 - 0.2189 = 1.4611 M

    [C] = 2X = 0.4378 M

    The equilibrium concentration of C is 0.4378 M
  2. 9 February, 04:43
    0
    [C] = 0.4248M

    Explanation:

    A + B ⇄ 2C

    C (i) 1.68M 1.68M 0.00

    ΔC - x - x + 2x

    C (eq) 1.68-x 1.68-x 2x

    Keq = [C]²/[A][B] = (2x) ² / (1.68 - x) ² = 8.98 x 10⁻²

    Take SqrRt of both sides = > √ (2x) ² / (1.68 - x) ² = √8.98 x 10⁻²

    => 2x/1.68 - x = 0.2895

    => 2x = 0.2895 (1.68 - x)

    => 2x = 0.4863 - 0.2895x

    => 2x + 0.2895x = 0.4863

    => 2.2895x = 0.4863

    => x = 0.4863/2.2895 = 0.2124

    [C] = 2x = 2 (0.2124) M = 0.4248M in 'C'
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Consider the reaction below that has a Keq of 8.98 X 10-2. The initial concentration of A and B is 1.68 M. Calculate the equilibrium ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers