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Question

A mixture of H2 and I2 in molecular proportion of 2 : 3 was heated at 440C till the reaction H2+I2 2HI reached equilibrium state. Calculate the percentage of iodine converted into HI (KCat 440C is 0.02)


  1. 3.38%

  2. 4.38%

  3. 5.38%

  4. 6.38%


Solution

The correct option is C

5.38%


H2+I22HIInitial moles 230AtEqb2xV3xV2XV

KC=4x2(2x)(3x)=0.02

199x2+5x6=0

x = 0.1615

Out of 3 moles , 0.1615 moles of I2 is converted into HI\)

Percentage of I2 converted to

HI=0.1615×1003=5.38

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