Chemistry Labs

Problem 6

The equilibrium constant of the reaction HX2+IX2⇌2 HI\ce{H2 + I2 <=> 2HI} is Kc=70.0K_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 2 of 3: Part (a-ii): 2 : 1 feed
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4α2(2−α)(1−α)=70.0⇒66α2−210α+140=0⇒α=0.951  (95.1 %)\dfrac{4\alpha^2}{(2 - \alpha)(1 - \alpha)} = 70.0 \Rightarrow 66\alpha^2 - 210\alpha + 140 = 0 \Rightarrow \alpha = 0.951\;(95.1\ \%)
Analysis

With [HX2]=c(2−α)[\ce{H2}] = c(2 - \alpha), [IX2]=c(1−α)[\ce{I2}] = c(1 - \alpha) and [HI]=2cα[\ce{HI}] = 2c\alpha: solving the quadratic gives α=0.951\alpha = 0.951 (the root <1< 1), so doubling hydrogen drives conversion up from 80.7 % to 95.1 % (Le Chatelier's principle).