hydrogen chloride gas reacts with oxygen gas to yield chlorine gas and water vapor in an equilibrium reaction. an experiment was performed in a closed vessel starting with a mixture of 0.50 M HCl and 0.050 M O2. the amount of chlorine was monitored until no change was observed. the final concentration of the chlorine was 0.048 M chlorine.
a) prepare an ICE table to answer the following:
b) what is the Keq for this reaction?
c) what are the equilibrium concentrations of all substances in the reaction?
........4HCl + O2 ==> 2Cl2 + 2H2O
initial.0.05..0.05....0......0
change...-4x....-x....2x.....2x
equil.0.05-4x.0.05-x.0.048...2x
The ICE table gives you all of the info you need to answer b and c.
a) We can set up an ICE table to organize the information:
Reaction: HCl + 0.5O2 ⇌ Cl2 + H2O
Initial Concentration (M):
HCl: 0.50
O2: 0.050
Cl2: 0
H2O: 0
Change in Concentration (M):
HCl: -x
O2: -0.5x
Cl2: +x
H2O: +x
Equilibrium Concentration (M):
HCl: 0.50 - x
O2: 0.050 - 0.5x
Cl2: x
H2O: x
b) The equilibrium expression for this reaction can be written as follows:
Keq = [Cl2][H2O] / [HCl][O2]
Given that the final concentration of Cl2 is 0.048 M, and we can assume the equilibrium concentration of H2O is also 0.048 M (since producing Cl2 and H2O is a 1:1 mole ratio), we can substitute these values into the equation:
Keq = (0.048)(0.048) / (0.50 - x)(0.050 - 0.5x)
c) To calculate the equilibrium concentrations, we need to solve the equation for x.
Using the given information that no change was observed once equilibrium was reached, we know that the concentration of Cl2 at equilibrium is 0.048 M. Thus, setting Cl2 = 0.048 in the equation, we can solve for x:
0.048 = (0.048)(0.048) / (0.50 - x)(0.050 - 0.5x)
Solving this equation will give us the value of x, which represents the equilibrium concentration of Cl2 (and also H2O). We can then substitute this value into the ICE table to find the equilibrium concentrations of all substances.
a) To answer the questions, let's first set up an ICE table.
The balanced equation for the reaction is:
2 HCl(g) + O2(g) ⇌ 2 Cl2(g) + 2 H2O(g)
We are given the initial concentrations for HCl and O2. Let's assume x is the change in concentration for Cl2 and H2O. Initially, there is no Cl2 or H2O, so their concentrations are zero.
The ICE table will look like this:
2 HCl(g) + O2(g) ⇌ 2 Cl2(g) + 2 H2O(g)
Initial 0.50 M 0.050 M 0 M 0 M
Change -2x -x +2x +2x
Equilibrium 0.50 - 2x 0.050 - x 2x 2x
b) To find the Keq for this reaction, we need to determine the equilibrium concentrations of all the substances. From the ICE table above, we can see that the equilibrium concentration of Cl2 is 0.048 M.
Keq = [Cl2]^2 / ([HCl]^2 * [O2])
Plugging in the given values, we have:
Keq = (0.048)^2 / ((0.50 - 2x)^2 * (0.050 - x))
Since no change was observed in the concentration of Cl2, we can assume that 2x = 0.048 M. Therefore, x = 0.024 M.
Now we can substitute the value of x into the equation:
Keq = (0.048)^2 / ((0.50 - 2(0.024))^2 * (0.050 - 0.024))
Keq = (0.048)^2 / (0.452^2 * 0.026)
Simplifying, we get:
Keq ≈ 5.38
c) Finally, let's find the equilibrium concentrations of all substances. From the ICE table:
[HCl] = 0.50 - 2x = 0.50 - 2(0.024) = 0.452 M
[O2] = 0.050 - x = 0.050 - 0.024 = 0.026 M
[Cl2] = 2x = 2(0.024) = 0.048 M
[H2O] = 2x = 2(0.024) = 0.048 M
So, the equilibrium concentrations are:
[HCl] ≈ 0.452 M
[O2] ≈ 0.026 M
[Cl2] ≈ 0.048 M
[H2O] ≈ 0.048 M