A mixture of 2.5 moles of H2O and 100 g of C are placed in a 50.0 L container and
allowed to come to equilibrium in the following reaction:
C(s) + H2O(g) CO(g) + H2(g)
The equilibrium concentration of hydrogen gas is found to be 0.040 M. What is the
equilibrium concentration of the water vapour?
(H2O) = 2.5 mol/50 L = 0.05 M
..........C + H2O ==> CO + H2
I.......solid.0.05.....0....0
C.......solid -x.......x....x
E......solid.0.05-x....x....x
Kc = (CO)(H2)/(H2O)
The problem give you H2, CO is the same H2, you plug in Kc and solve for (H2O)
You didn't give Kc in the problem and there isn't enough information to calculate it.
To find the equilibrium concentration of water vapor in this reaction, we need to use the stoichiometry and the given information.
First, let's determine the moles of C and H2O at the beginning:
- Moles of C (given mass of C): We can calculate the moles of carbon using its molar mass. The molar mass of carbon (C) is 12.01 g/mol. Therefore, moles of C = mass of C / molar mass of C = 100 g / 12.01 g/mol = 8.33 mol.
- Moles of H2O (given moles of H2O): The question states that 2.5 moles of H2O are present.
Next, let's see the balanced equation to determine the stoichiometry (molar ratio) between H2O and H2, as well as between H2O and CO.
C(s) + H2O(g) ⇌ CO(g) + H2(g)
From the equation, we can see that the molar ratio of water to hydrogen gas (H2) is 1:1. This means that for every mole of water that reacts, one mole of hydrogen gas is produced.
Now, let's use the molar ratio to find the equilibrium concentration of water vapor:
Given:
Initial moles of water (H2O) = 2.5 mol
Equilibrium concentration of hydrogen gas (H2) = 0.040 M
Since the molar ratio of H2O to H2 is 1:1, the concentration of water vapor would be the same as the concentration of hydrogen gas (H2). Therefore, the equilibrium concentration of water vapor is 0.040 M.
Thus, the equilibrium concentration of water vapor is 0.040 M.