Consider the following reaction:
2H2S+SO2=S(s)+H2O
A reaction mixture initially containing 0.500M H2S and 0.500M SO2 was found to contain 1.0×10−3M at a certain temperature. A second reaction mixture at the same temperature initially contains 0.255M H2S and 0.320M SO2. Calculate the equilibrium concentration of H2O in the second mixture at this temperature
write an expression for ka hor the dissociation of HBr, when HBr is dissociated, it loses its H+
Why did the H2O go to therapy? Because it wasn't feeling stable in the reaction mixture!
To calculate the equilibrium concentration of H2O in the second mixture, we need to use the stoichiometry of the reaction and the equilibrium constant. The equilibrium constant (K) for this reaction is given by K = [S(s)]/[H2S]^2[SO2].
Let's call the equilibrium concentration of H2O in the second mixture x. To determine x, we can set up an ice table and use the given initial concentrations and equilibrium concentrations:
Initial concentrations:
[H2S] = 0.255 M
[SO2] = 0.320 M
[S(s)] = 0 M (since it hasn't formed yet)
[H2O] = 0 M (since there is no H2O initially)
Change in concentration:
[H2S] = -x
[SO2] = -x
[S(s)] = +x
[H2O] = +x
Equilibrium concentrations:
[H2S] = 0.255 - x
[SO2] = 0.320 - x
[S(s)] = x
[H2O] = x
Now we can plug these concentrations into the equilibrium constant expression and solve for x:
K = [S(s)]/[H2S]^2[SO2]
1.0x10^-3 = x / (0.255 - x)^2(0.320 - x)
Now it's time for a bit of math magic to solve for x! Use your favorite algebraic techniques to rearrange the equation and solve for x.
After you've solved for x, you can substitute that value back into the expression for [H2O] to find the equilibrium concentration of H2O. Voila!
To determine the equilibrium concentration of H2O in the second mixture, we can use the stoichiometry of the reaction and the initial concentrations of H2S and SO2 in the second mixture.
First, let's write down the balanced equation for the reaction:
2H2S + SO2 → S(s) + 2H2O
From the stoichiometry, we can see that for every 2 moles of H2S reacted, 2 moles of H2O are formed.
Next, we'll use the initial concentrations of H2S and SO2 in the second mixture to calculate the change in concentration (Δ[ ]) for each reactant.
Initial concentration of H2S = 0.255 M
Change in concentration of H2S = -x (assuming x is the moles of H2S reacted)
Initial concentration of SO2 = 0.320 M
Change in concentration of SO2 = -x (assuming x is the moles of SO2 reacted)
Since the stoichiometry ratio is 1:1 for both H2S and SO2, the change in concentration of H2O will be 2x.
Now, let's calculate the equilibrium concentrations of H2S, SO2, and H2O. Since we know that the equilibrium concentration of H2O is 1.0 × 10^−3 M, we can set up an equation to solve for x:
2x = 1.0 × 10^−3 M
Solving for x, we get:
x = (1.0 × 10^−3 M) / 2 = 5.0 × 10^−4 M
Now that we have the value of x, we can substitute it back into the equations for the change in concentration:
Change in concentration of H2S = -x = -(5.0 × 10^−4 M) = -0.0005 M
Change in concentration of SO2 = -x = -(5.0 × 10^−4 M) = -0.0005 M
To find the equilibrium concentration of H2O, since 2 moles of H2O are formed for every 2 moles of the reactants, we can use the initial concentration of H2O in the second mixture (given as 0 M):
Equilibrium concentration of H2O = Initial concentration of H2O + 2x
= 0 M + 2(5.0 × 10^−4 M)
= 1.0 × 10^−3 M
Therefore, the equilibrium concentration of H2O in the second mixture at this temperature is 1.0 × 10^−3 M.
To calculate the equilibrium concentration of H2O in the second mixture, we need to first determine the change in concentration of H2S and SO2 from the initial concentrations to the equilibrium concentrations. Then, we will use the stoichiometry of the balanced chemical equation to find the change in concentration of H2O.
Let's denote the change in concentration of H2S as x and the change in concentration of SO2 as y.
According to the balanced chemical equation: 2H2S + SO2 = S(s) + H2O
The stoichiometric coefficient of H2S is 2, which means that for every 2 moles of H2S that react, 1 mole of H2O is produced.
Using this information, we can set up an ICE table for the reaction:
| 2H2S | + | SO2 | = | S(s) | + | H2O |
Initial (M) | 0.255 | + | 0.320 | = | 0 | + | 0 |
Change (M) | -x | - | y | = | 0.255 - x | + | 0 - x/2 |
At equilibrium, the concentrations of H2S and SO2 will be equal to their initial concentrations minus the change in concentration.
Now, let's set up the equilibrium expression for the reaction:
Kc = [S(s)] * [H2O] / [H2S]^2 * [SO2]
Given that the equilibrium concentration of H2O is 1.0×10−3M, we can substitute the initial concentrations and the change in concentration into the equilibrium expression:
1.0×10−3 = (0.255 - x/2) * x / (0.255 - x)^2 * (0.320 - y)
Since the value of x and y are small (compared to the initial concentrations of H2S and SO2), we can make the approximation that (0.255 - x) ≈ 0.255 and (0.320 - y) ≈ 0.320:
1.0×10−3 ≈ (0.255 - x/2) * x / (0.255)^2 * (0.320)
Now we can solve for x:
1.0×10−3 ≈ (0.255 - x/2) * x / (0.0650256)
Rearranging the equation:
(0.255 - x/2) * x ≈ 1.0×10−3 * (0.0650256)
0.255x - (x^2)/2 ≈ 6.50256×10−5
Rearrange further to form a quadratic equation:
0.5x^2 - 0.255x + 6.50256×10−5 ≈ 0
Solve this quadratic equation to find the value of x. Once you have determined x, substitute it back into the expression for the change in concentration of H2O:
Change in concentration of H2O = 0 - x/2
From this, you can calculate the equilibrium concentration of H2O by adding the change in concentration to the initial concentration of H2O.
Is that 1.0 x 10^-3 M H2O or something different. You should read your posts to make sure all of the information is there. I will assume it is (H2O). Have you checked the equation? It isn't balanced. I will balance it. And I wonder if the temperature is high enough so H2O is a gas? I will assume it is. In fact, I will write it as such in the equation.
2H2S + SO2 ==>3S(s) + 2H2O(g)
K = (H2O)^2/(H2S)^2(SO2).
Make an ICE chart.
Initial:
H2S = 0.5 M
SO2 = 0.5 M
H2O = 0
Change:
H2O = +1.0 x 10^-3
SO2 = -0.5 x 10^-3
H2S = -1.0 x 10^-3
Equilibrium:
H2S = 0.5-1.0 x 10^-3
SO2 = 0.5- 0.5) x 10^-3
H2O = 0 + 1.0 x 10^-3
Plug those equilibrium concns into the Keq expression I wrote above to find Keq.
Then prepare another ICE chart for the new mixture which will be
initial:
H2S = 0.255 M
SO2 = 0.320 M
H2O = 0
Change:
H2O = +2x
SO2 = -x
H2S = -2x
Equilibrium:
H2O = 2x
SO2 = 0.320- x
H2S = 0.255-2x
Plug those values into the Keq expression and using Keq you calculated before, determine the value of x. That times 2 will be the equilibrium value for H2O. Post your work if you get stuck.