Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: arnyk on May 09, 2006, 05:05:20 PM
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Looking over this multiple choice chem exam, couple are stumping me:
Ok here we go, this one's been bugging me:
Decomposition of CH3CHO to CH4.
r = k [CH3CHO]
k = 0.035 min-1
How long does it take for [CH3CHO]] to decrease from 0.4 M to 0.3 M?
A less than 1 minute
B 4 minutes
C 8 minutes
D 16 minutes
E greater than 16 minutes
I'm still messing with the units. r = mol / L min
If k is in min-1, *what* min-1? Is it mol / min? grams / minute? Need a little nudge here.
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This one looks like Hess:
If
2N2(g) + O2(g) <--> 2N2O(g) Kp = 5.04 x10-37
N2(g) + O2(g) <--> 2NO2(g) Kp = 4.23 x10-31
What is the Kp for-
2N2O(g) + O2(g) <--> 4NO(g)
Ok, what I did was "reverse" the first equation which gives you -5.04 x10-37.
Multiply equation 2 by "2" in order to get 4NO2(g) which gives you +8.43 x10-31.
Adding them cancels out into the final equation, and also gets you a final Kp of 8.46 x10-31.
The right answer is Kp = 3.55 x10-25.
My first thought is that you cannot do Hess's Law with K values? What then am I doing wrong?
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Looking over this multiple choice chem exam, couple are stumping me:
Balance the following chemical equation:
x Cr2O7-2 + y Cl- --> m Cr3+ + n Cl2
When it's balanced, what is the ratio x : y ?
A 1:1
B 1:2
C 1:3
D 2:3
E 1:6
The problem I'm having I suspect is that I do not fully understand the background knowledge required. The way I see it is that the charges are presently balanced. However, how can you balance the "O" when it is only on the left side? <-- That's the part you gotta whack into me for me to get it. The answer is 1:6 by the way.
In my opinion, the easiest way is writing the two half reactions, which often contain protons and molecules of water (generally speaking): of course, you just have a basic and incomplete chemical equation.
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The problem I'm having I suspect is that I do not fully understand the background knowledge required. The way I see it is that the charges are presently balanced. However, how can you balance the "O" when it is only on the left side? <-- That's the part you gotta whack into me for me to get it. The answer is 1:6 by the way.
To balance such equation you should take into account fact that it goes only in acidic conditions, thus you should add H+ on the left and H2O on the right.
But to give x:y answer you don't need to balance the equation - you just compare oxidation numbers.
Read more about balancing redox reactions (http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-redox) to get the idea.
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To balance such equation you should take into account fact that it goes only in acidic conditions, thus you should add H+ on the left and H2O on the right.
using this method you should end up with:
14H+ + Cr2O72- + 6Cl- --> 2Cr3+ + 3Cl2 + 7H2O
start with a 7 on the water to balance the oxygens, then a 14 on the H+ to balance the hydrogens, then 2 on the Cr3+ to balance Cr, then 6 on Cl- to balance charge, then 3 on Cl2 to balance Cl
But to give x:y answer you don't need to balance the equation - you just compare oxidation numbers.
This method also works. On the left side of the equation, Cr has an oxidation number of 6+ and Cl has an oxidation number of 1-. Therefore you would need 1 Cr to balance 6 Cl, hence the answer 1:6.
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using this method you should end up with:
14H+ + Cr2O72- + 6H2O --> 2Cr3+ + 3Cl2 + 7H2O
You have eaten chlorides :)
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Ok I see it now. Cr on the right side has an oxidation number of 6+. Since it has to gain 3 electrons to become Cr3+ on the product side...actually it has to gain 6 electrons since it is Cr2 meaning there must be 2 Cr3+ on the right. Those 6 electrons must come from Cl- (the reducing agent), therefore the coefficient on Cl- is 6.
Stay tuned for a few more guys! :D
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Ok here we go, this one's been bugging me:
Decomposition of CH3CHO to CH4.
r = k [CH3CHO]
k = 0.035 min-1
How long does it take for [CH3CHO]] to decrease from 0.4 M to 0.3 M?
A less than 1 minute
B 4 minutes
C 8 minutes
D 16 minutes
E greater than 16 minutes
I'm still messing with the units. r = mol / L min
If k is in min-1, *what* min-1? Is it mol / min? grams / minute? Need a little nudge here.
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min-1 is just 1/min. Multiply that by the concentration and you get mol/L min as your unit.
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How do you solve for minutes though?
If k is just the inverse of time, then it could be:
r = [CH3CHO] / t
System of equations?
r = k [CH3CHO]
k [CH3CHO] = [CH3CHO] / t
t = 1 / k
D'oh, back to where I started. ???
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Since your reaction is a first-order type (r=k[X]), you can use the integrated first-order rate law to solve the problem.
ln [new concentration] = -kt + ln [original concentration]
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Therefore,
t =[ ln[new] - ln[old] ] / (-k)
t = 8.22 minutes.
The answer is C 8 minutes. Correct! Thanks!
Now for the final:
This one looks like Hess:
If
2N2(g) + O2(g) <--> 2N2O(g) Kp = 5.04 x10-37
N2(g) + O2(g) <--> 2NO2(g) Kp = 4.23 x10-31
What is the Kp for-
2N2O(g) + O2(g) <--> 4NO(g)
Ok, what I did was "reverse" the first equation which gives you -5.04 x10-37.
Multiply equation 2 by "2" in order to get 4NO2(g) which gives you +8.43 x10-31.
Adding them cancels out into the final equation, and also gets you a final Kp of 8.46 x10-31.
The right answer is Kp = 3.55 x10-25.
My first thought is that you cannot do Hess's Law with K values? What then am I doing wrong?
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using this method you should end up with:
14H+ + Cr2O72- + 6H2O --> 2Cr3+ + 3Cl2 + 7H2O
You have eaten chlorides :)
oops, that first H2O was supposed to be a Cl-... fixed now
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You have to treat those Kp values exponentially.
When you reverse the reaction, the Kp is now a reciprocal. When you double everything in a reaction equation, the Kp is squared. The Kp for the overall reaction in your case is
(Kp1)-1(Kp2)2.