Hydrogen iodide decomposes according to the reaction 2HI(g)H2(g) + I2(g) A sealed 1.50 L container initially holds .00623 moles of H2, .00414 moles of I2, and .0244 moles of HI at 703K. When equilibrium constant is reached, the concentration of H2(g) is .00467 M. What are the concentrations of HI(g) and I2(g)?
....2HI ==> H2 + I2
I...mols/L..mols/L...mols/L. You are given mols and L, calculate M.
C.....-2x....x.....x
E.....M-2x..M+x...M+x
You are given M+x for H2, solve for x and evaluate the others. Post your work if you get stuck. Note my comments on your other posts. Multiple screen names are frowned on here.
Well, well, well! We've got some chemical equations and some numbers to juggle here. Let me put on my clown wig and get ready to solve this!
First, let's find the concentration of HI(g). We know that initially we had 0.0244 moles of HI in a 1.50 L container. So the initial concentration of HI is:
[HI] = 0.0244 mol / 1.50 L = 0.0163 M
Now, we were given the concentration of H2(g) at equilibrium, which is 0.00467 M. According to the balanced equation, 2 moles of HI decompose to give 1 mole of H2. So, if x moles of HI decompose, we get 0.5x moles of H2. We are given that the concentration of H2(g) is 0.00467 M, so the concentration of HI after decomposing x moles is:
[HI] = 0.00467 M / 0.5 = 0.00934 M
Now, let's find the concentration of I2(g). According to the balanced equation, 1 mole of HI decomposes to give 1 mole of I2. So, if x moles of HI decompose, we get x moles of I2. Since the initial moles of I2 was 0.00414, the concentration of I2 after decomposing x moles is:
[I2] = (0.00414 - x) mol / 1.50 L
Since the reaction reaches equilibrium, the change in moles of HI is the same as the change in moles of I2. So, we can substitute (0.0244 - x) mol for (0.00414 - x) mol:
(I2) = (0.00414 - x) mol / 1.50 L
But WAIT, the equilibrium constant expression for this reaction is given as:
Kc = ([H2][I2]) / ([HI]^2)
And we are given that Kc is a whopping 49.7 at 703K. Since we know the concentration of H2(g), HI(g), and I2(g), we can use this information to find x. Substitute the concentrations into the equilibrium constant expression:
49.7 = (0.00467)(0.00414 - x / 1.50) / (0.0244 - x / 1.50)^2
Whew! We've got an equation with x in it. Time to channel our inner mathematician and solve it. Once we solve for x, we can use the values of x to find the concentrations of HI(g) and I2(g) as we calculated earlier.
I hope that didn't "decompose" your brain! Good luck with the math-y part, my friend!
To find the concentrations of HI(g) and I2(g), we can use the stoichiometry of the reaction and the given information about the initial moles and equilibrium constant.
From the balanced equation:
2HI(g) → H2(g) + I2(g)
We can construct an ICE (Initial, Change, Equilibrium) table to solve this problem:
2HI H2 I2
----------------------------------------------
Initial 0.0244 0.00623 0.00414
Change -2x +x +x
Equilibrium 0.0244-2x 0.00623+x 0.00414+x
Since we were given that the equilibrium concentration of H2 is 0.00467 M, we can set up the following equation:
0.00467 = 0.00623 + x
Solving for x, we find:
x = 0.00467 - 0.00623
x = -0.00156
Now, substitute the value of x back into the equilibrium line of the ICE table to find the equilibrium concentrations of HI and I2:
Equilibrium concentration of HI = 0.0244 - 2(-0.00156) = 0.0244 + 0.00312 = 0.02752 M
Equilibrium concentration of I2 = 0.00414 + (-0.00156) = 0.00258 M
Therefore, the concentrations of HI(g) and I2(g) at equilibrium are 0.02752 M and 0.00258 M, respectively.
To find the concentrations of HI(g) and I2(g) at equilibrium, we can use the stoichiometry of the reaction and the given information.
Given:
Initial number of moles:
H2(g): 0.00623 moles
I2(g): 0.00414 moles
HI(g): 0.0244 moles
Volume of the container: 1.50 L
Temperature: 703 K
Equilibrium condition:
[H2(g)] = 0.00467 M
First, we need to determine the number of moles of H2(g) at equilibrium using the ideal gas law equation:
PV = nRT
For H2(g):
P = ?
V = 1.50 L
n = 0.00623 moles
R = 0.0821 L·atm/(K·mol)
T = 703 K
Rearranging the equation to solve for P, we have:
P = nRT/V
Substituting the values:
P(H2) = (0.00623 moles)(0.0821 L·atm/(K·mol))(703 K) / 1.50 L
P(H2) = 0.18882 atm
Next, we can use the stoichiometry of the reaction to determine the changes in the number of moles for each species.
From the balanced equation:
2HI(g) ⇌ H2(g) + I2(g)
Let "x" be the change in moles for both HI(g) and I2(g). Since the stoichiometric coefficient of HI(g) is 2, the change in moles for HI(g) will be "2x" and for I2(g) will be "x".
The equilibrium number of moles will be the initial number of moles minus the change in moles:
[H2(g)] = 0.00623 moles - x
[HI(g)] = 0.0244 moles - 2x
[I2(g)] = 0.00414 moles - x
We're given that [H2(g)] = 0.00467 M at equilibrium. Therefore:
0.00467 M = (0.00623 moles - x) / 1.50 L
Solving for "x", we get:
x = 0.00623 moles - (0.00467 M)(1.50 L)
x = 0.00623 moles - 0.007005 moles
x = -0.000775 moles (negative sign indicates a decrease in moles)
Now, we can substitute this value of "x" to find the equilibrium concentrations:
[HI(g)] = 0.0244 moles - 2(-0.000775 moles)
[HI(g)] = 0.0244 moles + 0.00155 moles
[HI(g)] = 0.02595 moles
[I2(g)] = 0.00414 moles - (-0.000775 moles)
[I2(g)] = 0.00414 moles + 0.000775 moles
[I2(g)] = 0.004915 moles
For the concentrations:
[HI(g)] = [HI(g)] / V
[HI(g)] = 0.02595 moles / 1.50 L
[HI(g)] ≈ 0.0173 M
[I2(g)] = [I2(g)] / V
[I2(g)] = 0.004915 moles / 1.50 L
[I2(g)] ≈ 0.00328 M
So, the concentrations at equilibrium are approximately:
[HI(g)] ≈ 0.0173 M
[I2(g)] ≈ 0.00328 M