Topic: Partial Pressure of 2L of 20atm O2 and 4L of 10atm H2 (Read 4587 times)

VinnyCee · « **on:** March 22, 2006, 08:32:55 PM »

There are three tanks, all initially closed to each other:

1) 2 Liters, with 20 atmosphere O₂.

2) 1 Liter, with nothing in it.

3) 4 Liters, with 10 atmosphere H₂.

What is the new total pressure when the system is opened?

The temperature is below 300K and the O₂ and H₂ react.

My work so far:

I balanced the O₂ and H₂ reaction to 2H₂ + 2O₂ = 2H₂O + O₂

But that wouldn't really matter because you just add the pressures to get the total pressure, right? So 20 atm + 10 atm = 30 atm TOTAL

I don't think it's right, please help.

AWK · « **Reply #1 on:** March 23, 2006, 03:20:16 AM »

For stoichiometry of reaction you should calculate moles of H2 and O2

Hunt · « **Reply #2 on:** March 23, 2006, 08:58:39 PM »

You need to find the new partial pressures in the container.
V_T = 2 + 1 + 4 = 7 L

If no rxn occurs, and at const temp:

P_f-Hydrogen = P_i-HydrogenV_i-Hydogen / V_T

P_f-Oxygen = P_i-OxygenV_i-Oxygen / V_T

P_i-Hydrogen = 10 atm
P_i-Oxygen = 20 atm

V_i-Oxygen = 2 L
V_i-Hydrogen = 4 L

P_T = P_f-Oxygen + P_f-Hydrogen

If a rxn occurs, then , I assume the water formed is in the liquid state. The reaction would be a complete one, combustion of Hydrogen. However, to find the pressures, we need to know how much Oxygen and Hydrogen are present. If the Temp is given, we can determine the number of moles and proceed to calculate the final pressure.

2H₂_(g) + O₂_(g) ---> 2H₂O_(l)

Topic: Partial Pressure of 2L of 20atm O2 and 4L of 10atm H2 (Read 4587 times)

VinnyCee

Partial Pressure of 2L of 20atm O2 and 4L of 10atm H2

AWK

Re:Partial Pressure of 2L of 20atm O2 and 4L of 10atm H2

Hunt

Re:Partial Pressure of 2L of 20atm O2 and 4L of 10atm H2

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