For the equilibrium reaction:
CO(g)+H2O(g)<=>CO2(g)+H2(g)
the Keq value at 690°C is 10.0. A mixture of 0.300 mol of CO, 0.300 mol of H2O, 0.500 mol of CO2 and 0.500 mol of H2 is placed in a 1.0 L flask.
a)Write the Keq expression for this reaction and use the concentrations above to calculate a trial value of the Keq. Does this value match the given Keq ? Is the reaction at equilibrium ? Explain.
b)Calculate the equilibrium concentrations of all four species.
Keq = (CO2)(H2)/(CO)(H2O)
I assume by trial value of Keq you mean the reaction quotient. That sill be
Kquo = (0.500)(0.500)/(0.300)(0.300)
Kquo = 2.78. It doesn't match Keq of 10 and it is too low which means that the products are too small and the reactants are too large so the reaction will go to the right.
..........CO + H2O ==> CO2 + H2
initial.0.300.0.300...0.500.0.500
change...-x...-x........+x.....+x
equil..0.3-x..0.3-x..0.5+x..0.5+x
Substitute the equilibrium values of the ICE chart above and solve for x, then the individual concns of each molecule.
Post your work if you get stick. Here is a hint: When you set up the numbers in for Keq to solve the equation, note that you can take the square root of both sides and you need not solve a quadratic equation. I get x to be approximately 0.11 but that's approximate.
a) The Keq expression for the reaction is given by:
Keq = [CO2] * [H2] / [CO] * [H2O]
Using the given concentrations, we can substitute them into the expression:
Keq_trial = (0.500 mol/L) * (0.500 mol/L) / (0.300 mol/L) * (0.300 mol/L)
= 8.33 mol/L
The trial value of Keq is 8.33, which does not match the given Keq of 10.0. Therefore, the reaction is not at equilibrium.
b) To calculate the equilibrium concentrations, we'll need to use an ICE (Initial Change Equilibrium) table.
CO(g) + H2O(g) <=> CO2(g) + H2(g)
Initial: 0.300 0.300 0.500 0.500
Change: -x -x +x +x
Equilibrium: 0.300 - x 0.300 - x 0.500 + x 0.500 + x
Since the reaction has not reached equilibrium, we cannot directly calculate the equilibrium concentrations.
a) The Keq expression for the given reaction is:
Keq = [CO2] * [H2] / [CO] * [H2O]
Using the given concentrations:
[CO] = 0.300 mol / 1.0 L = 0.300 M
[H2O] = 0.300 mol / 1.0 L = 0.300 M
[CO2] = 0.500 mol / 1.0 L = 0.500 M
[H2] = 0.500 mol / 1.0 L = 0.500 M
Plugging in these values into the Keq expression:
Keq = (0.500 M) * (0.500 M) / (0.300 M) * (0.300 M)
= 0.417
The calculated trial Keq value is 0.417. This value does not match the given Keq value of 10.0. Therefore, the reaction is not at equilibrium.
b) To calculate the equilibrium concentrations, we need to use an ICE table. Let's assume the change in the concentration of CO2 and H2 is "x".
Initially, at equilibrium:
[CO] = 0.300 M
[H2O] = 0.300 M
[CO2] = 0.500 M
[H2] = 0.500 M
Using the ICE table:
CO(g) + H2O(g) <=> CO2(g) + H2(g)
Initial: 0.300 0.300 0.500 0.500
Change: -x -x +x +x
Equilibrium: 0.300-x 0.300-x 0.500+x 0.500+x
Since the number of moles of CO and H2O is equal and the number of moles of CO2 and H2 is also equal, we can set up the following equation:
[x * (0.500+x)] / [(0.300-x)*(0.300-x)] = 10.0
Now we need to solve this equation to find the value of x. However, since the initial concentrations of all species are equal, we can say that x is very small compared to the initial concentration. Therefore, we can approximate (0.300 - x) as 0.300 and (0.500 + x) as 0.500 in the denominator.
[(x * (0.500+x)] / [(0.300)*(0.300)] = 10.0
Simplifying the equation,
(0.500+x)^2 = 10.0 * (0.300)^2
0.250 + x + x^2 = 0.270
Rearranging terms,
x^2 + x - 0.020 = 0
Solving this quadratic equation using the quadratic formula, we find that x ≈ 0.137 or x ≈ -1.137.
Since negative values do not make sense in this context, we can discard the negative value. Therefore, x ≈ 0.137.
Substituting this value back into the equilibrium concentrations:
[CO] = 0.300 - x = 0.300 - 0.137 = 0.163 M
[H2O] = 0.300 - x = 0.300 - 0.137 = 0.163 M
[CO2] = 0.500 + x = 0.500 + 0.137 = 0.637 M
[H2] = 0.500 + x = 0.500 + 0.137 = 0.637 M
Therefore, the equilibrium concentrations are approximately:
[CO] = [H2O] ≈ 0.163 M
[CO2] = [H2] ≈ 0.637 M
a) The Keq expression for this reaction is given by the ratio of the products' concentrations to the reactants' concentrations, each raised to the power of their respective stoichiometric coefficients:
Keq = ( [CO2]^a * [H2]^b ) / ( [CO]^c * [H2O]^d )
Where a, b, c, and d are the stoichiometric coefficients of CO2, H2, CO, and H2O, respectively.
To calculate a trial value of Keq, we need to substitute the given concentrations into the Keq expression:
[CO2] = 0.500 mol/1.0 L = 0.500 M
[H2] = 0.500 mol/1.0 L = 0.500 M
[CO] = 0.300 mol/1.0 L = 0.300 M
[H2O] = 0.300 mol/1.0 L = 0.300 M
Plugging these values into the Keq expression:
Keq_trial = (0.500^1 * 0.500^1) / (0.300^1 * 0.300^1)
= 0.25 / 0.09
= 2.778
The trial value of Keq, 2.778, does not match the given Keq of 10.0. Therefore, the reaction is not at equilibrium.
b) To calculate the equilibrium concentrations of all four species, we need to set up an ICE table to track the change in concentration for each species during the reaction. The ICE table stands for Initial, Change, Equilibrium.
Initial:
[CO2] = 0.500 M
[H2] = 0.500 M
[CO] = 0.300 M
[H2O] = 0.300 M
Change:
Let the change in concentration for each species be represented by x.
[CO2] = 0.500 - x
[H2] = 0.500 - x
[CO] = 0.300 + x
[H2O] = 0.300 + x
Equilibrium:
At equilibrium, the concentrations will be equal to the initial concentrations plus the change.
Now, we can substitute these equilibrium concentrations into the Keq expression and solve for x. The equation becomes:
10.0 = ([CO2] * [H2]) / ([CO] * [H2O])
Substituting the equilibrium concentrations:
10.0 = ((0.500 - x) * (0.500 - x)) / ((0.300 + x) * (0.300 + x))
Solve this equation to find the value of x, and then use it to calculate the equilibrium concentrations of all four species.