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November 08, 2025, 04:07:50 am

Author Topic: Keq q  (Read 930 times)  Share 

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vamsiaus

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Keq q
« on: July 18, 2009, 02:11:11 pm »
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This is an easy question but i am getting a different answer to that in the solutions, so just thought I would ask:
A 2.00 mol sample of nitrosyl bromide, NOBr, was introduced into a 5.0 L vessel. At equilibrium, the concetration of nitrosyl bromide was found to be .0750 M. Calculate the equilibrium constant for the reaction shown in the equation:
        
                                                           2NOBr(g)   (equilibrium Arrow)    2NO(g)  +  Br2 (g)
What I already did was to work out the equation to go ((1.625 ^2) x (.8125)) / (.375^2) which gives me 15.26, but the answer is 3 something. so just wondering if somebody can help me out with what i have done wrong. this question was part of a worksheet our teacher gave us as a preparation for the upcoming sac. So yeah Thanks to anybody who can help.
« Last Edit: July 18, 2009, 02:21:38 pm by vamsiaus »
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monokekie

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Re: Keq q
« Reply #1 on: July 18, 2009, 02:33:05 pm »
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A 2.00 mol sample of nitrosyl bromide, NOBr, was introduced into a 5.0 L vessel. At equilibrium, the concetration of nitrosyl bromide was found to be .0750 M. Calculate the equilibrium constant for the reaction shown in the equation:
2NOBr(g)   (equilibrium Arrow)    2NO(g)  +  Br2 (g)

kk,
n(NOBr)in equilibrium= cxv=0.075x5=0.375
n(NO)produced=n(NOBr)consumed=n(NOBr)initial-n(NOBr)in equilibrium=2-0.375=1.625
n(Brs)produced=1/2n(NO)produced=0.8125
c(NOBr)=0.075, c(NO)=1.625/5=0.325,C(Br2)=0.8125/5=0.1625

0.325^2 x 0.1625 / 0.075^2 gives 3.05 =]

vamsiaus

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Re: Keq q
« Reply #2 on: July 18, 2009, 02:39:57 pm »
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Omg i forgot to divide the number of mole of NOBr by volume to get concentration of .075 for the last step. thanks for your help Monokekie
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vamsiaus

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Re: Keq q
« Reply #3 on: July 18, 2009, 02:57:08 pm »
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alright another q from the same worksheet.
Water undergoes self ionistion according to the following equation.
2H20 (l)  (eq. arrow) H30+ (l) + OH- (aq)   delta h > 0
The pH of pure wter at 10 degress is
 A. less than 7
B. greater than 7
C. Equal to 7
D. Greater thanor less than 7 depending upon the source of the water

If someone could explain why one answer is the right one, it would be greatly appreciated.
Thanks.
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TrueTears

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Re: Keq q
« Reply #4 on: July 18, 2009, 03:09:19 pm »
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The pH of water at 25 degrees is 7.

As the temp decreases the back reaction is preferred. [As it releases heat]

Now that means less products are formed. Less is formed. Hence the pH increases [getting more basic] Hence it is B
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vamsiaus

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Re: Keq q
« Reply #5 on: July 18, 2009, 03:14:04 pm »
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Thanks TT
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