keq=(2x)^2/ (.06-x)(.03-x)=33
4x^2=33(.0018+x^2-.09x)
-29x^2+.09&33 x- .0018*33=0
ok,use the quadratic eqauation, solve for x.
2x is the conc of HI
.06-x is the conc of H2
.03-x is the conc of I2
at some temperature, Keq=33 for the reaction H2 + I2 -> 2HI. If initially, [H2]= .0600 M and [I2]= .0300 M, what are all three equilibrium concentrations?
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