The equilibrium: quinone (Q) + leuco dye (L) ⇌ hydroquinone (H) + coloured dye (D), may be studied conveniently by observing the colour of the dye D. Two millimole of D was added to 1.00 L of solution

Question

The equilibrium: quinone (Q) + leuco dye (L) ⇌ hydroquinone (H) + coloured dye (D), may be studied conveniently by observing the colour of the dye D. Two millimole of D was added to 1.00 L of solution that was 0.30 M in H and 0.015 M in Q. It was then found that 5.0% of the added D was reduced to L. What is the equilibrium constant for the above reaction? The equation is balanced with 1 molecule of each of the four substances, and the change in the concentrations of H and Q may be neglected.

Answer:

380.00

Solution:

Dye D added = 2 mmol in 1.00 L = 2 × 10^-3 M D reduced = 5.0% of 2 × 10^-3 = 1 × 10^-4 M

[L] = 1 × 10^-4 M and [D] = 1.9 × 10^-3 M

[H] ≈ 0.30 M and [Q] ≈ 0.015 M

K = [H][D]/([Q][L])

K = (0.30)(1.9 × 10^-3)/[(0.015)(1 × 10^-4)]

K = 380, written as 0380

TestHub · Accepted Answer

Dye D added = 2 mmol in 1.00 L = 2 × 10 -3 M D reduced = 5.0% of 2 × 10 -3 = 1 × 10 -4 M [L] = 1 × 10 -4 M and [D] = 1.9 × 10 -3 M [H] ≈ 0.30 M and [Q] ≈ 0.015 M K = [H][D]/([Q][L]) K = (0.30)(1.9 × 10 -3 )/[(0.015)(1 × 10 -4 )] K = 380, written as 0380

Chemistry - Chemical Equilibrium Question with Solution | TestHub

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