The following reaction occurs in a 2.0L reactor at 800 degrees c.

CO+H20-CO2+H2
The reactor is charged initially with 5.0 atm of CO and 8.0 atm of water vapor. Calculate the final pressure of hydrogen in the reactor.

The solution to this is not obvious to me without a Kp or Kc

To calculate the final pressure of hydrogen in the reactor, we need to determine the stoichiometry of the reaction and then apply the ideal gas law.

First, let's determine the balanced equation for the reaction:

CO + H2O -> CO2 + H2

The stoichiometry of the reaction tells us that for every 1 molecule of CO, we produce 1 molecule of H2. Therefore, the pressure of H2 formed will be directly proportional to the pressure of CO consumed.

Next, let's determine the mole ratio between CO and H2. Since both gases are at the same temperature and pressure, we can use the ideal gas law to calculate their initial and final moles.

For CO:
nCO = (P x V) / (R x T) = (5.0 atm x 2.0 L) / (0.0821 atm L/mol K x 800 K) ≈ 0.123 mol

For H2O:
nH2O = (P x V) / (R x T) = (8.0 atm x 2.0 L) / (0.0821 atm L/mol K x 800 K) ≈ 0.32 mol

Since the stoichiometry ratio between CO and H2 is 1:1, the moles of H2 formed will also be 0.123 mol.

Finally, we can calculate the final pressure of H2 using the ideal gas law:

PH2 = (nH2 x R x T) / V = (0.123 mol x 0.0821 atm L/mol K x 800 K) / 2.0 L ≈ 3.13 atm

Therefore, the final pressure of hydrogen in the reactor is approximately 3.13 atm.