Assume that gaseous hydrogen iodide is synthesized from hydrogen gas and iodide vapor at 50 degrees celcius where the equilibrium constant is 1.00 x 10^2. Suppose HI at 5.00 10^-1 M, H2 at 1.00 x 10^-2 M, and I2 at 5.00 x 10^-3 M are mixed in a 5.0 L container. Calculate all concentrations at equilibruim.
I have made an assumption that the numbers listed at the beginning of the experiment, are in fact, molarity and not moles.
H2 + I2 ==> 2HI
Set up ICE chart, substitute into Keq expression, and solve for x, then calculate the concn of each component.
You may want to calculate Qeq first (I obtained 5,000 for that) which tells us that HI is too high, the others are too low; therefore, the reaction must move to the left.
initial concn:
H2 = 0.01 M
I2 = 0.005 M
HI = 0.5 M
change in concn:
H2 = +x
I2 = +x
HI = -2x
equilibrium concn:
H2 = 0.01+x
I2 = 0.005+x
HI = 0.5-2x
By the way, note the correct spelling of celsius.
This problem shifts right by definiition of a Kp that is > 1
To calculate the concentrations at equilibrium, we will use the given values and the equilibrium constant (K) for the reaction.
First, let's write the balanced equation for the reaction:
H2 + I2 ⇌ 2HI
Next, we need to create an ICE (Initial, Change, Equilibrium) table to track the concentrations of the species involved in the reaction.
Initial concentrations:
[H2] = 1.00 x 10^-2 M
[I2] = 5.00 x 10^-3 M
[HI] = 5.00 x 10^-1 M
Let's assume that the concentration of HI at equilibrium is x M. Therefore, the change in concentration for H2 and I2 will be -x (since they are reactants being consumed) and the change for HI will be +2x (since it is being produced).
ICE table:
H2 + I2 ⇌ 2HI
Initial: 1.00 x 10^-2 5.00 x 10^-3 5.00 x 10^-1
Change: -x -x +2x
Equilibrium: 1.00 x 10^-2 - x 5.00 x 10^-3 - x 5.00 x 10^-1 + 2x
Now, we can use the equilibrium constant expression to relate the concentrations at equilibrium:
K = [HI]^2 / [H2]·[I2]
Substituting the equilibrium concentrations into the equation:
1.00 x 10^2 = (5.00 x 10^-1 + 2x)^2 / (1.00 x 10^-2 - x)(5.00 x 10^-3 - x)
By solving this equation, we can find the value of x, which represents the concentration of HI at equilibrium. However, this equation is not easily solved by hand, so it requires numerical methods or a calculator.
Assuming you have access to a calculator or a computer program, you can solve the equation to find x. Once you have x, you can calculate the concentrations of H2 and I2 using the equilibrium expressions:
[H2] = 1.00 x 10^-2 - x
[I2] = 5.00 x 10^-3 - x
These values represent the equilibrium concentrations of H2, I2, and HI in the 5.0 L container at 50 degrees Celsius.