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Elden [556K]
3 years ago
6

At equilibrium at 1200°C, [I2] = 9.5 ×10–2 M and [I] = 3.2 × 10–2 M. What is the value of Keq for this system?

Chemistry
1 answer:
Lelu [443]3 years ago
3 0

Answer:The value of equilibrium constant is 1.0778\times 10^{-2}.

Explanation:

I_2\rightleftharpoons 2I^-

[I_2]=9.5\times 10^{-2} M,[I^-]=3.2\times 10^{-2} M

Expression of an equilibrium constant is given as:

K_{eq}=\frac{[I^-]^2}{[I_2]}=\frac{3.2\times 10^{-2}\times 3.2\times 10^{-2}}{9.5\times 10^{-2}}

K_{eq}=1.0778\times 10^{-2}

The value of equilibrium constant is 1.0778\times 10^{-2}.

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