At equilibrium, a mixture contains [H$_2$] = 0.2 mol dm$^{-3}$, [I$_2$] = 0.2 mol dm$^{-3}$ and [HI] = 1.6 mol dm$^{-3}$ for the reaction H$_2$ + I$_2$ ⇌ 2HI. Calculate $K_c$.
A0.0156
B64CORRECT
C8
D40
JUPEBChemistryChemical Equilibrium
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$K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \dfrac{(1.6)^2}{0.2 \times 0.2} = \dfrac{2.56}{0.04} = 64$. Forgetting to square [HI] gives 40, and inverting the expression gives 0.0156.
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