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Question
At 700 K, the equilibrium constant for the reaction
\[\ce{H_{2(g)} + I_{2(g)} ↔ 2HI_{(g)}}\]
is 54.8. If 0.5 molL–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?
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Solution
It is given that equilibrium constant `"K"_"C"` for the reaction
\[\ce{H_{2(g)} + I_{2(g)} ↔ 2HI_{(g)}}\] is 54.8.
Therefore, at equilibrium, the equilibrium constant `"K'"_"C"` for the reaction
\[\ce{2HI_{(g)} ↔ H_{2(g)} + I_{2(g)}}\] will be `1/54.8`
[HI] = 0.5 `" mol L"^(-1)`
Let the concentrations of hydrogen and iodine at equilibrium be x mol L–1
`["H"_2] = ["I"_2] = x " mol" " L"^(-1)`
Therefore, `(["H"_2]["I"_2])/["HI"]^2 = "K'"_"C"`
`=> (x xx x)/(0.5)^2 = 1/54.8`
`=> x^2 = 0.25/54.8`
`=> x = 0.06754`
x = 0.068 `" mol L"^(-1)` (approximatley)
Hence, at equilibrium, `["H"_2] = ["I"_2] = 0.068 " mol L"^(-1)`
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