Hydrogen gas and iodine gas react via the equation, H2 + I2 <--> 2HI, and K=76 at 600K. If 0.05 mole of HI is placed in a flask at 600K, what are the equilibrium concentrations of HI, I2, H2?

I am very lost at where to even start with this problem

Are you sure there is no volume given. The following will give you mols at equilibrium.

..........2HI ==> H2 + I2
initial...0.05.....0....0
change....=2x......x.....x
equil...0.05-x.....x.....x

Kc = 76 = (H2)(I2)/(HI)^2
Solve for x
Then x/volume = concn.

I think you mean 0.05 - 2x for the change of (2HI)

Well, let me break it down for you in a more relatable way.

Think of the reaction between hydrogen gas (H2) and iodine gas (I2) as a chaotic dance party. The equation H2 + I2 <--> 2HI represents the confused mingling and exchanging of partners between the molecules.

Now, at the start of the party, you have some guests already on the dance floor. In this case, you've got 0.05 moles of HI in the flask, who are already twisting and twirling away.

But remember, this is an equilibrium situation, so the dance party is ongoing. Some of the HI molecules will collide and break apart, releasing hydrogen (H2) and iodine (I2) into the mix.

To figure out the equilibrium concentrations, we use the equilibrium constant (K), which is given as 76. Think of K like the popularity rating of the dance moves. The higher the K, the more popular the moves are.

To find the equilibrium concentrations, we'll need to set up an ICE table. ICE stands for Initial, Change, and Equilibrium.

Start by writing down the initial concentration of HI, I2, and H2. In this case, the initial concentration of HI is 0.05 moles.

Next, we have to figure out the changes in the concentrations. Since the balanced equation shows that 1 mole of HI produces 1 mole of I2 and 1 mole of H2, we can assume that the change in the concentration of I2 and H2 is the same.

Finally, we'll add up the initial concentration and the change in concentration to get the equilibrium concentration.

However, to find the change in concentration, we need to know the initial concentrations of I2 and H2, which are not given in the problem. Without those initial values, it's impossible to determine the equilibrium concentrations.

So, let's go back to the dance party analogy. If we don't even know how many I2 and H2 molecules were present initially, it's like trying to guess how many people are dancing at the party without knowing how many people arrived in the first place.

Long story short, without the initial concentrations of I2 and H2, we can't determine the equilibrium concentrations of HI, I2, and H2. Hope that clears things up, or at least brings a smile to your face!

To solve this problem, we need to use the equation and the given value of K to set up an expression for the equilibrium concentrations of HI, I2, and H2.

First, let's define the initial concentration of HI, which is 0.05 moles, as [HI]initial = 0.05 M.

The reaction equation suggests that for every 1 mole of HI that reacts, 1 mole of H2 and 1 mole of I2 are produced. Therefore, initially we have [H2]initial = [I2]initial = 0 M.

Now, let's assume that at equilibrium, the changes in the concentrations are x mol/L for both H2 and I2, and 2x mol/L for HI. This is because the coefficients in the balanced equation suggest that the change in concentration of HI is twice the changes in concentrations of H2 and I2.

Now we can write the equilibrium concentrations as:
[HI] = (0.05 + 2x) M
[H2] = x M
[I2] = x M

Now, let's use the expression for K to set up an equation:

K = ([H2] * [I2]) / [HI]^2

Substituting the values of the concentrations, we get:
76 = (x * x) / (0.05 + 2x)^2

Solving this equation will give us the value of x, which represents the change in concentration.

From here, we need to substitute the value of x back into the expressions for the equilibrium concentrations to find the final values.

I hope this helps! Let me know if you have any further questions.

To solve this problem, you can use the concept of equilibrium constants and the stoichiometry of the chemical equation. Here's a step-by-step guide to help you solve it:

1. Start by writing down the balanced chemical equation: H2 + I2 ⇌ 2HI.

2. Define the equilibrium concentrations of each species:
Let [H2], [I2], and [HI] represent the equilibrium concentrations of hydrogen gas, iodine gas, and hydrogen iodide (HI), respectively.

3. Use the given information that 0.05 mole of HI is placed in the flask. Here, the initial concentration of HI is equal to its equilibrium concentration, assuming no H2 or I2 initially.

4. Since the coefficients of H2 and I2 in the balanced equation are both 1, and the coefficient of HI is 2, you can express the equilibrium concentrations of H2 and I2 as [H2] = [I2] = x.

5. For the concentration of HI, you know that [HI] = 0.05 (given in the problem).

6. Now, using the equilibrium constant expression, K = [HI]^2/([H2] * [I2]), substitute the expressions for each concentration:
K = (0.05)^2 / (x * x)

7. The value of the equilibrium constant K is given as 76 at 600K, so you can substitute this value into the equation:
76 = (0.05)^2 / (x * x)

8. Rearrange the equation and solve for x:
x^2 = (0.05)^2 / 76
x^2 = 0.000003289
x ≈ 0.001810 (rounded to four decimal places)

9. Since the concentration of H2, I2, and HI are all equal to x in this case, you can use this value to find the equilibrium concentrations:
[H2] = x ≈ 0.0018
[I2] = x ≈ 0.0018
[HI] = 0.05 (given in the problem)

Therefore, at equilibrium, the concentrations of H2 and I2 are approximately 0.0018 M, and the concentration of HI is 0.05 M.