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    Working with Kp for the equilibrium

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    At an unspecified temperature, the equilibrium reaction system I2(g) + Cl2(g) <--> 2ICl(g) has a partial pressure of ICl(g) that is 5520 times the partial pressure of I2(g). In turn, the partial pressure of I2(g) is 2.75 times the partial pressure of Cl2(g). What is Kp for the equilibrium at this unspecified temperature?

    © BrainMass Inc. brainmass.com December 24, 2021, 4:53 pm ad1c9bdddf
    https://brainmass.com/chemistry/chemical-equilibrium/working-kp-equilibrium-14080

    SOLUTION This solution is FREE courtesy of BrainMass!

    Partial pressure of ICl = 5520* pI2

    Partial Pressure of I2, pI2 = 2.75 * pCl2

    therefore pICl = 5520*2.75* pCl2 = 15180 pCl2

    We have, for a reaction of the form, mA+nB &lt;--&gt; xC + yD,

    Kp = {(pC)^x * (pD)^y}/(pA)^m * (pB)^n

    where pC is the partial pressure of C and so on.

    Here Kp = (pICl)^2/(pI2 * pCl2) = (15180*pCl2)^2/(2.75 * pCl2)* (pCl2)

    = 15180^2/2.75 = 83793600

    Kp for the forward reaction is very high. (Please verify whether the given number is 5520 or 5.520)

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 4:53 pm ad1c9bdddf>
    https://brainmass.com/chemistry/chemical-equilibrium/working-kp-equilibrium-14080

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