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The equilibrium constant for the system below at 1 000 K is 3.5 ´ pr007-1.jpg.
I2(g) Û 2I(g)
If 1.00 mol of iodine gas, I2, is placed in a 2-L flask at 1 000 K, what will be the concentrations of I2 and I when the system reaches equilibrium?
I2(g) Û 2I(g)
If 1.00 mol of iodine gas, I2, is placed in a 2-L flask at 1 000 K, what will be the concentrations of I2 and I when the system reaches equilibrium?
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GPT-5 mini
Kc = 3.5 × 10^-3 for I2(g) ⇌ 2 I(g).
Initial: [I2]0 = 1.00 mol / 2.00 L = 0.500 M, [I]0 = 0.
Let x = [I]eq. Then [I2]eq = 0.500 − x/2, and
Kc = [I]^2/[I2] = x^2/(0.500 − x/2) = 3.5×10^-3.
Solve: x^2 = 1.75×10^-3(1 − x) → x^2 + 1.75×10^-3 x − 1.75×10^-3 = 0.
Positive root: x = 0.0410 M.
Thus [I]eq = 4.10×10^-2 M and [I2]eq = 0.500 − 0.0205 = 0.480 M.
(As moles in the 2.00‑L flask: n(I) ≈ 0.0820 mol, n(I2) ≈ 0.960 mol.)
Initial: [I2]0 = 1.00 mol / 2.00 L = 0.500 M, [I]0 = 0.
Let x = [I]eq. Then [I2]eq = 0.500 − x/2, and
Kc = [I]^2/[I2] = x^2/(0.500 − x/2) = 3.5×10^-3.
Solve: x^2 = 1.75×10^-3(1 − x) → x^2 + 1.75×10^-3 x − 1.75×10^-3 = 0.
Positive root: x = 0.0410 M.
Thus [I]eq = 4.10×10^-2 M and [I2]eq = 0.500 − 0.0205 = 0.480 M.
(As moles in the 2.00‑L flask: n(I) ≈ 0.0820 mol, n(I2) ≈ 0.960 mol.)
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