[arnyk]

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.

--------------

This one looks like Hess:

If

2N_2(g) + O_2(g) <--> 2N₂O_(g)   Kp = 5.04 x10^-37
N_2(g) + O_2(g) <--> 2NO_2(g)   Kp = 4.23 x10^-31

What is the Kp for-

2N₂O_(g) + O_2(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 4NO_2(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?

[Albert]

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.

[Borek]

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 to get the idea.

[syko sykes]

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⁺ + Cr₂O₇^2- + 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.

[Borek]

using this method you should end up with:
 14H⁺ + Cr₂O₇^2- + 6H₂O --> 2Cr³⁺ + 3Cl₂ + 7H₂O

You have eaten chlorides

[arnyk]

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! 

[arnyk]

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.

[wereworm73]

min^-1 is just 1/min.  Multiply that by the concentration and you get mol/L min as your unit.

[arnyk]

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. 

[wereworm73]

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]





   



 

[arnyk]

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

2N_2(g) + O_2(g) <--> 2N₂O_(g)   Kp = 5.04 x10^-37
N_2(g) + O_2(g) <--> 2NO_2(g)   Kp = 4.23 x10^-31

What is the Kp for-

2N₂O_(g) + O_2(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 4NO_2(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?

[syko sykes]

using this method you should end up with:
 14H⁺ + Cr₂O₇^2- + 6H₂O --> 2Cr³⁺ + 3Cl₂ + 7H₂O

You have eaten chlorides

oops, that first H₂O was supposed to be a Cl^-... fixed now

[wereworm73]

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.