For the reaction H2+I22HI Kc is 49. Calculate the concentration of HI at equilibrium when initially one mole of H2 is mixed with one mole of I2 in 2 litre flask.?
1 Answer
Explanation:
We're asked to find the equilibrium concentration of
The equilibrium constant expression for this reaction is
The initial concentrations of both
So our initial concentrations for each species are
Initial:

#"H"_2# :#0.5M# 
#"I"_2# :#0.5M# 
#"HI"# :#0#
From the coefficients of the chemical equation, we can predict the changes in concentration with the quantity
Change:

#"H"_2# :#x# 
#"I"_2# :#x# 
#"HI"# :#+2x#
and the final concentrations are the sum of the initial and change:
Final:

#"H"_2# :#0.5Mx# 
#"I"_2# :#0.5Mx# 
#"HI"# :#2x#
It's algebra time! Let's plug these into our equilibrium constant expression to start solving for
We can neglect the solution that is greater than
Therefore, the final equilibrium concentration of