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