Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

SO3(g) decomposes according to the equation 2SO3(g) → 2SO2(g) + O2(g) A sealed container contains 0.50 mol of SO3 gas at 100°C and 2 atmospheres pressure. What would be the pressure in the container if the SO3 gas is decomposed completely according to the above equation and the temperature were maintained at 100°C ?

Chemistry
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

Because its a sealed container its volume is constant + and temperature is also constant: PV = nRT => V/RT = contstant too.. 2SO3 => 2SO2 + O2 0.5 mole 0.5 mole 0.25 mole V/RT = constant = n(initial)/P(initial) = n(final)/P(final) n(inital) = n(SO3) = 0.5 mole, P(initial) = 2atm, n(final) = n(SO2) + n(O2) => 0.5 mole + 0.25 mole = 0.75 mole, P(final) = x atm n(inital)/P(initial) = 0.5 mole/2 atm = 0.75 mole / x atm => 0.5x = 1.5 => x = 3 atm = P(final)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question