................H2 + I2 ==> 2HI
initial.........x.....x......0
equilibrium....0.080..0.060...0.490
Use these equilibra values to calculate Kc for the reaction. It isn't given in the problem. Then the second part of the problem takes off from the first.
..................H2 + I2 ==> 2HI
initial.........0.080..0.060...0.790
change...........+x....+x.....-2x
equilibrium...0.080+x..0.060+x...0.790-x
Substitute into the Kc expression and solve for x.
Post your work if you get stuck. Note:The 0.790 for HI initially comes from the 0.490 at equilibrium + 0.300M = 0.790M
Some hydrogen and iodine are mixed up at 229 degrees celsiusin a 1-L container. When equilibrium is established, the following concentrations are present: [HI]=
0.490M, [H2]=0.080M, and [I2]= 0.060M. If an additional 0.300 mol of HI is then added, what concentrations of all species will be present when the equilibrium is established?
4 answers
so its (.490-x)2/(.080+x)(.060+x)? cuz my teacher said something just goes away because it is so small so that is why I am confused
No, it is (0.790-2x)^2/(0.080+x)(0.060+x)
I didn't write the -2x in the chart but I should have done so. As to one of the numbers "going away" because it's so small, I don't know how you know that until you work the problem.And that means you solve this problem the hard way, with the quadratic formula but the calculator will make it easier. And there is no confusion. The chemistry part is over when you write the Kc expression. It's math the rest of the way.
I didn't write the -2x in the chart but I should have done so. As to one of the numbers "going away" because it's so small, I don't know how you know that until you work the problem.And that means you solve this problem the hard way, with the quadratic formula but the calculator will make it easier. And there is no confusion. The chemistry part is over when you write the Kc expression. It's math the rest of the way.
x_0.33