Asked 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.
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.
Answers
Answered by
bobpursley
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.
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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