Question

Two gases X and Y are found in the atmosphere in only trace amounts because they...

Two gases X and Y are found in the atmosphere in only trace amounts because they decompose quickly. When exposed to ultraviolet light the half-life of X is 0.75h, while that of Y is 90.min. Suppose an atmospheric scientist studying these decompositions fills a transparent 5.0L flask with X and Y and exposes the flask to UV light. Initially, the partial pressure of X is 3.0 times greater than the partial pressure of Y. Will the partial pressure of x ever be lower than y. If so at what time will it be lower?

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Half life (t1/2) is the time required by a substance to get decomposed or reduced to half of its original amount. Now here given

Half life (t1/2) for X = 0.75h = 45 min

Half life (t1/2) for Y = 90 min

That means half life of Y is greater than half life of X. So X will be reduces first or it will take just half the time of Y to get reduced.

Now given initially the partial pressure of X is 3.0 times greater than the partial pressure of Y. Now we know partial pressure of a gas is directly proportional to its number of moles. So we can say initially moles of X is 3 times greater than Y

So lets say we have initially 3x moles of X and x moles of Y

Now once they get start decomposing,

After 45 mins,

X will be reduced to 3x/2 = 1.5x moles

And Y will be reduced by 1/4 (because Y takes 90 min to reduced to 1/2. So in the 1st 45 min, only 14 of it will decomposed). So we can say, after 45 mins, moles of Y left = x-1/4x = 3/4x = 0.75‬ moles

So here after 45 mins, moles of Y is still less than moles of X.

So partial pressure of X is greater than partial pressure of Y

After 90 mins,

X will be reduced to = 1.5x *1/2 = 0.75X

And Y will be reduced to = 1/2 = 0.5 moles (as half life of Y is 90 min)

So here after 90 mins, moles of Y is still less than moles of X.

So partial pressure of X is greater than partial pressure of Y

Now at 180 mins,

amount of X left = 0.75x * 1/4 = 0.1875‬x

But moles of Y left = x/2 * 1/2 = 0.25x.

Now here mols of Y left is more than mols of X left.

So after 180 mins. partial pressure of X will be lower than partial pressure of Y

Add a comment
Know the answer?
Add Answer to:
Two gases X and Y are found in the atmosphere in only trace amounts because they...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • Two gases X and Y are found in the atmosphere only trace amounts because they decompose quickly. When exposed to ultrav...

    Two gases X and Y are found in the atmosphere only trace amounts because they decompose quickly. When exposed to ultraviolet light the half-life of Xis 0.50 h, while that of Y is 90. min. Suppose an atmospheric scientist studying these decompositions fills a transparent 5.0 L flask with Xand Y and exposes the flask to UV light. Initially, the partial pressure of X is 3.0 times greater than the partial pressure of Y. dlo yes As both gases decompose,...

  • Two gases X and Y are found in the atmosphere in only trace amounts because they...

    Two gases X and Y are found in the atmosphere in only trace amounts because they decompose quickly. When exposed to ultraviolet light the half-life of X is 75. min, while that of Y is 2.25 h. Suppose an atmospheric scientist studying these decompositions fills a transparent 10.0 L flask with X and Y and exposes the flask to UV light. Initially, the partial pressure of X is 25.0% greater than the partial pressure of Y. yes As both gases...

  • PLEASE ANSWER ALL OF THEM!! What happens to the partial pressures of two ideal gases, X...

    PLEASE ANSWER ALL OF THEM!! What happens to the partial pressures of two ideal gases, X and Y, when an inert gas is added to this mixture in a closed metal container? (A) both decrease (B) both increase (C) one increases, the other decreases (D) both stay the same When 3.84 L of hydrogen react according to the reaction below, what volume (in L) of oxygen is needed? Assume the reaction occurs at constant temperature and pressure. 2H2(g) + O2(g)...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT