That's 0.15 WHAT and 0.05 WHAT? You add whatever it is. It is either mols, molar, or atm pressure. I will assume it is M since you say "concentrations." but you can change that as needed.
..........H2 + I2 ==> 2HI
I......0.15M...0.05M...0
C........-x.....-x.....2x
E......0.15-x...0.05-x..2x
Keq = 4.5E-6 = (HI)^2/(H2)(I2)
Substitute from the ICE chart and solve for x and the concns of each. Post your work if you get stuck.
For the reaction H2(g) + I2(g) ¡ê 2 HI(g), you have the initial concentrations [H2] = 0.15 and [I2] = 0.05. Keq for the reaction at this temperature is 4.5 x 10-6. Make a reaction table. Include rows for initial concentration, change in concentration, and equilibrium concentration. Write down the equation for Keq. Solve for x. What are the equilibrium concentrations for H2, I2, and HI?
2 answers
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2 answers