Hydrogen and carbon dioxide react at a high temperature to give water and carbon monoxide.
H2(g) + CO2(g) --> H2O(g) + CO(g)
(a) Laboratory measurements at 986°C show that there are 0.11 mol each of CO and H2O vapor and 0.087 mol each of H2 and CO2 at equilibrium in a 1.0 L container. Calculate the equilibrium constant for the reaction at 986°C.
I got 2.1e2. I did (.11)(.11)/(.087)(.087)
What are i doing wrong?
(b) Suppose 0.052 mol each of H2 and CO2 are placed in a 1.5 L container. When equilibrium is achieved at 986°C, what amounts of CO(g) and H2O(g), in moles, would be present? [Use the value of K calculated in part (a).]
[CO]
______ mol
[H2O]
______ mol
The numbers you have for Keq are correct; I assume you are punching the wrong buttons on your calculator. For (0.11)^2/(0.087)^2 I get 1.6
For part b.
0.052moles/1.5L = 0.0347M
..............H2 + CO2 ==> H2O + CO
initial...0.0347..0.0347.....0.....0
change......-x......-x.......+x....+x
equil...0.0347-x..0.0347-x....x.....x
Substitute into Ka expression and solve for x. The question asks for moles; therefore, x is molarity and M x L = moles.
Post your work if you get stuck.
thanks
Would you get the same answer for both parts of b?
Since you started with the same molarity for each and you used x of each, then the moles of each will be the same at equilibrium.
ok so I did (x)(x)/(.03466)(.03466)= 1.6. I solved for x and got .0438. Do I have to do anything with this number to get the correct answer?
Can you check my work?
Thanks:)
I would have rounded the 0.034666 to 0.03467 (and that's too many significant figures). However, it's (x)(x)/(0.03467-x)^2. What did you do with the -x in the 0.03467 - x term? Throw it away. You can't do that (at least not in this problem). You must solve the quadratic. I don't get anything like 0.0438M for an answer. After you find the M for x, you must subtract it from 0.03467 to find (CO2) and (H2), then multiply (CO2), (H2), (H2O), and (CO) by 1.5 to obtain moles (since the problem asks for each of these in moles.)
Sorry I am soo confused right now O.O. is there anyway u can explain this in simplier form?
To calculate the equilibrium constant for the reaction at 986°C, we need to use the values of the equilibrium concentrations given to us. The equilibrium constant, K, is calculated as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients.
(a) To calculate the equilibrium constant, we can use the balanced chemical equation and the given equilibrium concentrations:
H2(g) + CO2(g) --> H2O(g) + CO(g)
From the given information, at equilibrium:
[CO] = 0.11 mol/L
[H2O] = 0.11 mol/L
[H2] = 0.087 mol/L
[CO2] = 0.087 mol/L
Note that the concentrations are in mol/L because we are dealing with a 1.0 L container.
Now we can use these concentrations to calculate the equilibrium constant. The formula to calculate the equilibrium constant is: K = ([H2O] * [CO]) / ([H2] * [CO2])
K = (0.11 * 0.11) / (0.087 * 0.087) = 1.699 (approximately)
So, the equilibrium constant for the reaction at 986°C is approximately 1.699.
It seems like you made a mistake in your calculation by squaring the concentrations of both CO and H2O instead of multiplying them together. To correct it, use the equation K = (0.11 * 0.11) / (0.087 * 0.087) = 1.699 (approximately).
(b) To determine the amounts of CO(g) and H2O(g) at equilibrium when 0.052 mol each of H2 and CO2 are placed in a 1.5 L container, we can use the equilibrium constant (K) calculated in part (a).
Given:
[H2] = 0.052 mol/L
[CO2] = 0.052 mol/L
K = 1.699
The balanced chemical equation allows us to determine the stoichiometric coefficients:
H2(g) + CO2(g) --> H2O(g) + CO(g)
Using the stoichiometry of the reaction, we can express the concentrations of CO and H2O at equilibrium in terms of x (the change in concentration):
[CO]eq = x
[H2O]eq = x
Substituting these values into the expression for K, we get:
K = ([H2O]eq * [CO]eq) / ([H2] * [CO2]) = (x * x) / (0.052 * 0.052)
Now, rearrange the equation and solve for x:
x^2 = K * ([H2] * [CO2])
x^2 = 1.699 * (0.052 * 0.052)
x^2 = 0.0042143
Taking the square root of both sides:
x = √0.0042143
x ≈ 0.0649
So, at equilibrium, there would be approximately 0.0649 mol of CO(g) and 0.0649 mol of H2O(g) present in the 1.5 L container.
Thus, the amounts of CO(g) and H2O(g) at equilibrium would be approximately:
[CO] ≈ 0.0649 mol
[H2O] ≈ 0.0649 mol