Thursday

April 24, 2014

April 24, 2014

Posted by **Chris** on Sunday, January 20, 2013 at 8:54pm.

N2O5 to NO2 and O2

initial concentration is 1.65 X 10-2mol/L

rate constant is 4.80 X 10-4/s

time is 825s

According to my book, the answer is 1.24 X 10-2 mol/L and t= 4.80 X 10 (3)s

I have no idea how to set this problem up with the 1st order rate law...

- Chemistry -
**DrBob222**, Sunday, January 20, 2013 at 9:23pmThis is a first order reaction (you know that from the units given for k).

ln(No/N) = kt

ln(0.0165/N) = 4.80E-4*600

Solve for N.

Use the same equation. No is the same, N is 10% of that, k is the same, solve for t.

Check my work for typos.

- Chemistry -
**jacfar**, Sunday, January 20, 2013 at 11:01pmA vessel containing 39.5 cm3 of helium gas at 25°C and 106 kPa was inverted and placed in cold ethanol. As the gas contracted, ethanol was forced into the vessel to maintain the same pressure of helium. If this required 17.8 cm3 of ethanol, what was the final temperature of the helium?

- Chemistry -
**DrBob222**, Sunday, January 20, 2013 at 11:22pmHoe many mols He were in the container? That's PV = nRT. Solve for n. Use that n in another PV = nRT. You know P,

V = 39.5cc - 17.8cc

R and solve for T.

Rememberr V must be i L, R is 0.08206 if you use P in atm or 8.314 if you use P in kPa. T must be in kelvin.

- To jacfar -
**DrBob222**, Sunday, January 20, 2013 at 11:24pmYou should not piggy back your question on another post. Go to the top of the page and click on "Post a New Question."

- Chemistry -
**Chris**, Monday, January 21, 2013 at 8:53amSo, to solve for t would I set up the problem like this...

ln (0.00124/0.0165)=4.80 X 10-4 X t?

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