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Reaction H2 +I2 yields 2HI
All three gases are initially at 0.1atm/ upon reaching equilibrium it is found that H2 pressure droped by 55% what is the equilibrium constant for this reaction.

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    if pressure dropped by 55 percent, then the new concentration is 45 precent of the original. I assume you mean Partial pressure of the H2 dropped by 55 percent.
    Pressure is directly related to concentration.
    keq= [HI]^2/[H2][I2]

    keq=(.1+2x)^2/(.1-x)(.1-x)
    but .1-x= .55(.1) or x=.1-.055=.045
    which means
    keq=(.1+2*.045)^2/.055^2=11.9 check my math and thinking.

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