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?
................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
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.
x_0.33
To solve this problem, we need to use the concept of the equilibrium constant and the stoichiometry of the chemical reaction.
First, let's write down the balanced chemical equation for the reaction between hydrogen and iodine to form hydrogen iodide:
H2(g) + I2(g) ⇌ 2HI(g)
According to the given information, at equilibrium, the concentrations of HI, H2, and I2 are [HI] = 0.490 M, [H2] = 0.080 M, and [I2] = 0.060 M, respectively.
The equilibrium constant expression for this reaction is:
Kc = [HI]² / ([H2] * [I2])
Now, let's calculate the value of the equilibrium constant, Kc.
Using the given equilibrium concentrations, we can substitute the values into the expression for Kc:
Kc = (0.490 M)² / ((0.080 M) * (0.060 M))
= 3.43 / 0.0288
= 119.10 (rounded to two decimal places)
Now, let's determine the concentrations of all species after 0.300 mol of HI is added.
Initially, the concentration of HI before the addition is 0.490 M. Adding 0.300 mol of HI will increase the total amount of HI:
Total moles of HI = Initial moles of HI + Added moles of HI
= (0.490 M)(1 L) + 0.300 mol
= 0.490 mol + 0.300 mol
= 0.790 mol
The volume of the container is still 1 L, so the concentration of HI after the addition will be:
[HI] = Total moles of HI / Total volume
= 0.790 mol / 1 L
= 0.790 M (rounded to three decimal places)
Since the stoichiometry of the reaction is 1:1:2 for H2:I2:HI, the concentration of H2 and I2 can be calculated by using the equation:
[H2] = Initial concentration of H2 - Added concentration of H2
= 0.080 M - 0.300 M
= -0.220 M (negative because we subtracted)
[I2] = Initial concentration of I2 - Added concentration of I2
= 0.060 M - 0.300 M
= -0.240 M (negative because we subtracted)
However, concentrations cannot be negative, so the concentrations of H2 and I2 will be zero:
[H2] = 0 M
[I2] = 0 M
Therefore, when the equilibrium is established after adding 0.300 mol of HI, the concentrations of all species will be:
[HI] = 0.790 M
[H2] = 0 M
[I2] = 0 M