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When 15 mole of H2 is mixed with 5.2 mole of I2 at equilibrium 10 mole of HI is formed....

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lxsptter | Student, Undergraduate | (Level 2) Valedictorian

Posted July 1, 2012 at 5:51 PM via web

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When 15 mole of H2 is mixed with 5.2 mole of I2 at equilibrium 10 mole of HI is formed. What is the equilibrium constant for the reaction

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted July 1, 2012 at 6:04 PM (Answer #1)

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Hydrogen reacts with iodine to form hydrogen iodide in a reaction that has the following chemical reaction.

H2 + I2 --> 2HI

Initially there are 15 moles of hydrogen gas and 5.2 moles of iodine gas. There is no hydrogen iodide to begin with. After the reaction reaches equilibrium 10 mole of hydrogen iodide is formed.

The equilibrium constant of the reaction is given by: [HI]^2/[H2]*[I2]

= (10/V)^2/((10/V)*(0.2/V))

=> 100/(10*0.2)

=> 50

The equilibrium constant for the reaction is 50

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