Find the a)partial pressure of ethylene oxide and b)the total gas pressure, after 30 hours in a reaction vessel at 415C if the initial pressure of ethylene oxide is 782mmHg. k= 2.05x 10^-4 t(1/2)=56.3
Part A)
I converted hours to seconds, and used the formula ln(Pa)t = 108000s x k + ln782
I got the correct answer which is 1.9x10^-7mmHg.
But I don't know how to do part B
Well, part B seems like a bit of a pressure-filled question! But don't worry, I'm here to help make it a gas!
To find the total gas pressure, we need to consider all the gases present in the reaction vessel. The partial pressure of ethylene oxide that we found in part A is just one part of the puzzle.
To calculate the total gas pressure, we need to take into account the partial pressure of ethylene oxide (which we found to be 1.9x10^-7 mmHg) and any other gases that might be present. Unfortunately, you haven't provided any information about other gases in the reaction vessel, so I can't give you a specific answer.
However, if you know the partial pressures of all the other gases in the reaction vessel, you can simply add them up to get the total gas pressure.
Now, if you're feeling a bit uncertain about what other gases might be present, you could always assume that the reaction vessel is perfectly sealed and that no gas leaks out or gets in. In that case, the total gas pressure would remain constant at whatever value it was initially (assuming no changes in temperature or volume).
I hope this helps lighten the pressure of your question a bit!
To calculate the total gas pressure, you need to consider the partial pressure of ethylene oxide as well as any other gases present in the reaction vessel. If the partial pressure of ethylene oxide is denoted as P(EtO), then the total gas pressure (P(total)) can be obtained by summing up the partial pressures of all the gases present.
Since you are given the partial pressure of ethylene oxide (P(EtO)) = 1.9x10^-7 mmHg, and assuming no other gases are present, you can simply state:
Part B)
The total gas pressure after 30 hours in the reaction vessel will be equal to the partial pressure of ethylene oxide, which is 1.9x10^-7 mmHg.
In this case, the total gas pressure is equal to the partial pressure of the only gas present, which is ethylene oxide.
To find the total gas pressure, you need to consider the reaction that is taking place in the reaction vessel. If the reaction is only producing or consuming ethylene oxide gas, then the total gas pressure would be the sum of the partial pressure of ethylene oxide and any other gases present.
However, since you haven't mentioned the specific reaction or any other gases, I'll assume that the reaction is occurring in a closed system with no other gases involved. In this case, the total gas pressure will remain constant throughout the reaction, and it will be equal to the initial pressure of ethylene oxide, which is 782 mmHg.
So, the answer to part B) is 782 mmHg.