Set up an ICE chart.
............H2 + I2 ==> 2HI
initial.0.240...0.240....0
change......-x....-x.....2x
equil...0.240-x..0.240-x..2x
Substitute into Kc expression.
Kc = 54.3 = (HI)^2/(H2)(I2)
Solve for x and evaluate 0.240-x
Kc for the reaction of hydrogen and iodine to produce hydrogen iodide,
H2(g)+I2(g)f 2HI(g)
is 54.3 at 430C. What will the concentrations be at equilibrium if we start with 0.240M concentrations of both H2 and I2?
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