Problem 6

The equilibrium constant of the reaction

\ce{H2 + I2 <=> 2HI}

K_c = 70.0

at 600 °C. (a) What percentage of iodine is converted at equilibrium if the reactants are mixed at 600 °C in (i) a 1 : 1 molar ratio, and (ii) a 2 : 1 molar ratio (twice as much hydrogen as iodine)? (b) How many moles of hydrogen must be mixed with 1 mole of iodine so that 99 % of the iodine is converted to hydrogen iodide at equilibrium at 600 °C?

Step 1 of 3: Part (a-i): Equimolar feed

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K = \dfrac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]} = \dfrac{4\alpha^2}{(1 - \alpha)^2} = 70.0 \Rightarrow \dfrac{2\alpha}{1 - \alpha} = \sqrt{70} \approx 8.3666 \Rightarrow \alpha = 0.807\;(80.7\ \%)

Analysis

Taking the square root of both sides avoids a quadratic equation: $2\alpha = 8.3666(1 - \alpha) \Rightarrow 10.3666\,\alpha = 8.3666 \Rightarrow \alpha = 0.807$ , or 80.7 % conversion of iodine.