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JEE Main 2016 · Chemistry · Chemical Equilibrium · medium

00 mol of H2 and 2.00 mol of I2 are heated in a 1.00 L closed vessel until H2(g)+I2(g)rightleftharpoons2HI(g) reaches equilibrium, where Kc=50. The…

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1.00 mol1.00\ \mathrm{mol} of H2\mathrm{H_2} and 2.00 mol2.00\ \mathrm{mol} of I2\mathrm{I_2} are heated in a 1.00 L1.00\ \mathrm{L} closed vessel until H2(g)+I2(g)⇌2HI(g)\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)} reaches equilibrium, where Kc=50K_c=50. The amount of HI\mathrm{HI} at equilibrium is nearly
  1. A. 1.50 mol1.50\ \mathrm{mol}
  2. B. 0.93 mol0.93\ \mathrm{mol}
  3. C. 1.87 mol1.87\ \mathrm{mol}
  4. D. 2.00 mol2.00\ \mathrm{mol}
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Answer: (C) 1.87 mol1.87\ \mathrm{mol}

Let xx mol of H2\mathrm{H_2} react, so [HI]=2x[\mathrm{HI}]=2x, [H2]=1−x[\mathrm{H_2}]=1-x, [I2]=2−x[\mathrm{I_2}]=2-x (volume is 11 L). Kc=(2x)2(1−x)(2−x)=50K_c=\dfrac{(2x)^2}{(1-x)(2-x)}=50, giving 4x2=50(2−3x+x2)4x^2=50(2-3x+x^2), i.e. 46x2−150x+100=046x^2-150x+100=0. The root below 11 is x=150−22500−1840092=0.934x=\dfrac{150-\sqrt{22500-18400}}{92}=0.934. Hence HI=2x=1.87\mathrm{HI}=2x=1.87 mol. The root x=2.33x=2.33 is rejected because it exceeds the available H2\mathrm{H_2}. The value 0.930.93 is xx itself, not 2x2x.
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