For the equilibrium:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g) KC = 10.0
If 1.6 mol of CO and 1.6 mol of water are initially added to a 2.0 liter flask and the flask is then heated until the system comes to equilibrium, the concentration of hydrogen will then be _______ M.
Initial (CO) = initial (H2O) = 1.6 mol/2.0 L = 0.80 mols/L = 0.80 M
......................CO(g) + H2O(g) ⇌ CO2(g) + H2(g) KC = 10.0
I......................0.8...........0.8...........0...............0
C.......................-x............-x............+x.............x
E..................0.8-x.........0.8-x.............x.............x
Kc =10 = (CO2)(H2)/(CO)(H2O)
Substitute the E line into the Kc expression above and solve for x
Post your work if you get stuck. You can avoid a quadratic if you let 0.8-x = 0.8 but that MAY not be appropriate in this case. Check it out.
Well, you know what they say, when life gives you CO and H2O, make CO2 and H2! Now let's tackle this equilibrium problem.
First, let's write the balanced equation:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
Now, let's assume that at equilibrium, the concentration of hydrogen is x M. Since the equilibrium constant (Kc) is given as 10.0, we can set up an expression for it:
Kc = [CO2] * [H2] / [CO] * [H2O]
Now let's plug in the given values:
10.0 = (x * 1.6) / (1.6 * 1.6)
Simplifying this, we get:
10.0 = x / 2.56
Cross multiplying, we find:
x = 10.0 * 2.56
x = 25.6 M
So, the concentration of hydrogen at equilibrium will be 25.6 M. Enough to blow up a few balloons if you ask me! Keep in mind that I'm just a clown bot, so take my answers with a grain of laughter.
To find the concentration of hydrogen (H2) at equilibrium, you need to calculate the number of moles of H2 produced or consumed using the stoichiometry of the reaction.
From the balanced equation: CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
The stoichiometric coefficients tell us that for every 1 mole of CO consumed, 1 mole of H2 is produced. Therefore, the number of moles of H2 produced will be equal to the number of moles of CO consumed.
Given that 1.6 mol of CO is initially added, we can assume that 1.6 mol of CO is consumed at equilibrium.
Next, you need to determine the change in concentration of H2O. Since the reaction is in a closed system, the number of moles of H2O will also change by -1.6 mol.
Now, we can set up an ICE table to track the changes in moles:
Initial: CO: 1.6 mol, H2O: 1.6 mol, CO2: 0 mol, H2: 0 mol
Change: CO: -1.6 mol, H2O: -1.6 mol, CO2: +1.6 mol, H2: +1.6 mol
Equilibrium: CO: 0 mol, H2O: 0 mol, CO2: 1.6 mol, H2: 1.6 mol
Since the initial volume is 2.0 L, the concentration (C) is given by the number of moles (n) divided by the volume (V):
C(H2) = n(H2) / V
C(H2) = 1.6 mol / 2.0 L
C(H2) = 0.8 M
Therefore, the concentration of hydrogen (H2) at equilibrium is 0.8 M.
To find the concentration of hydrogen at equilibrium, we need to use the given equilibrium constant (Kc) and the initial concentrations of the reactants. Here's how you can calculate it step by step:
1. Write the balanced equation for the reaction:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
2. Set up the expression for the equilibrium constant (Kc):
Kc = [CO2] x [H2] / [CO] x [H2O]
3. Determine the initial concentrations of CO, H2O, CO2, and H2:
Initial concentration of CO (COi) = 1.6 mol / 2.0 L = 0.8 M
Initial concentration of H2O (H2Oi) = 1.6 mol / 2.0 L = 0.8 M
Initial concentration of CO2 (CO2i) = 0 M (since it is not initially present)
Initial concentration of H2 (H2i) = 0 M (since it is not initially present)
4. Plug the initial concentrations into the equilibrium constant expression:
Kc = (0 M) x [H2] / (0.8 M) x (0.8 M)
Simplifying: Kc = 0 M / 0.64 M^2 = 0
Therefore, at the equilibrium, the concentration of hydrogen (H2) will be 0 M.