One quick chem question :)

[Swenis]

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Hey guys, I recently came across this problem that I have done, but I seem to be off by a tenth. Can anyone tell me what I've done wrong? First, the question:

"At 1285 oC, the equilibrium constant for the reaction Br2(g) <-> 2Br(g) is Keq=0.133. A 0.200-L vessel containing an equilibrium mixture of the gases has 0.245g Br2(g) in it. What is the mass of the Br(g) in the vessel?"

Ok, here is what I've done so far to get to the end of the question:

1) Determined the number of moles of Br2 gas in equilibrium then plugged all of the numbers into PV=nRT and solved for the pressure, 0.980 atm.

2) Then, I solved the expression Keq = P(Br)^2 / P(Br2). 0.133 = P(Br)^2 / .980. P(Br) = .361 atm.

Ok, the book confirmed both of the values given above for pressures, here is where is get mixed up.

P=nRT/V
(.361)(0.200)=n(.0821)(1558)
5.645E-4 = n
(5.645E-4 moles Br(g))(79.904g Br(g)/1 mole Br(g)) = .0451 😕

The book says the answer is .451 yet I only managed .0451. Can anyone please figure out where I went wrong? The campus is shut down for a bit and I can't get to the solutions. 😍

 

[CuRy]

2) Then, I solved the expression Keq = P(Br)^2 / P(Br2). 0.133 = P(Br)^2 / .980. P(Br) = .361 atm.

Ok, the book confirmed both of the values given above for pressures, here is where is get mixed up.

P=nRT/V
(.361)(0.200)=n(.0821)(1558)
5.645E-4 = n
(5.645E-4 moles Br(g))(79.904g Br(g)/1 mole Br(g)) = .0451 😕

The book says the answer is .451 yet I only managed .0451. Can anyone please figure out where I went wrong? The campus is shut down for a bit and I can't get to the solutions. 😍

I double-checked! Everything is making sense to me. It's very hard to come an exact number, except the place of decimal. So, you are good to go.