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 3 of 3: Part (b): Moles of H2 for 99 % conversion

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\dfrac{(1.98)^2}{(x - 0.99)(0.01)} = 70.0 \Rightarrow x - 0.99 = \dfrac{3.9204}{0.70} = 5.60 \Rightarrow x = 6.59\ \text{mol}

Analysis

For 1 mole of I2 with conversion 0.99, at equilibrium $[\ce{I2}] = 0.01$ , $[\ce{HI}] = 1.98$ , and $[\ce{H2}] = x - 0.99$ . Substituting into Kc gives $x - 0.99 = 5.60$ , so $x = 6.59$ moles of $\ce{H2}$ are required.