should i ever use ICE tables on the real test?

[tdod]

---

Members do not see this ad.

Or is it too time consuming?

 

[pepocho]

.

 

[orangetea]

noooooooo. you should memorize the rule:
If Keq is < 10^3 and all your reactants are present(initially) shift is negligible.
If Keq > 10^3 and all your products are present (initially) shift is negligible
If Keq=1 and your products/reactants are present(initially) shift is negligible


The PS section imo is shady math and test taking skills 😀

 

[sillyjoe]

noooooooo. you should memorize the rule:
If Keq is < 10^3 and all your reactants are present(initially) shift is negligible.
If Keq > 10^3 and all your products are present (initially) shift is negligible
If Keq=1 and your products/reactants are present(initially) shift is negligible


The PS section imo is shady math and test taking skills 😀

Can you explain this further please?

 

[justadream]

TBR has a shortcut (pH = 1/2pKa - 1/2[HA]) that allows you to bypass ICE charts for many problems. However, I don't know how applicable it is to most MCAT testing situations.

I know there are restrictions for when this is useful but does anyone know whether the MCAT regularly goes beyond those restrictions?

 

[orangetea]

Can you explain this further please?

It's in the TBR equilibrium chem chapter.

So if you have let's say ( I am using their example)

N2 + 3 H2 --> 2NH3

Kp= 3.0 x 10^-5
Partial Pressure:
N2= 3.75 atm
H2= 2.0 atm

What is the partial pressure of NH3 once equilibrium has reached assuming there is no ammonia in the system initially.
IF you were to set up an ice table it would be

Initially: 3.75 2.00 0
Shift: -x -3x +2x
Equi: 3.75-x 2.00-x 2x

3.0x10^-5= (2x)^2/(3.75-x)(2.00-x)^3

HOWEVER you see that your Keq is 10^-5 which is <<<<< 10^-3 meaning that most of your reactants shifted so if (x or 3x shifts) it's not going to make much of a difference anyway. So

3.0x10^-5= (2x)^2/(3.75)(2.00)^3 which makes your math easier and will give you the correct answer 🙂

 

[orangetea]

TBR has a shortcut (pH = 1/2pKa - 1/2[HA]) that allows you to bypass ICE charts for many problems. However, I don't know how applicable it is to most MCAT testing situations.

I know there are restrictions for when this is useful but does anyone know whether the MCAT regularly goes beyond those restrictions?

This is just another way of calculating pH for weak acids. Every problem I have done using this equation has worked. You have to think realistically there is about 1.4 min/question. The math/problems on the exam are indicative of that.

 

[sillyjoe]

It's in the TBR equilibrium chem chapter.

So if you have let's say ( I am using their example)

N2 + 3 H2 --> 2NH3

Kp= 3.0 x 10^-5
Partial Pressure:
N2= 3.75 atm
H2= 2.0 atm

What is the partial pressure of NH3 once equilibrium has reached assuming there is no ammonia in the system initially.
IF you were to set up an ice table it would be

Initially: 3.75 2.00 0
Shift: -x -3x +2x
Equi: 3.75-x 2.00-x 2x

3.0x10^-5= (2x)^2/(3.75-x)(2.00-x)^3

HOWEVER you see that your Keq is 10^-5 which is <<<<< 10^-3 meaning that most of your reactants shifted so if (x or 3x shifts) it's not going to make much of a difference anyway. So

3.0x10^-5= (2x)^2/(3.75)(2.00)^3 which makes your math easier and will give you the correct answer 🙂

I thought you can drop the X's because most of the reactants didn't shift to products?

 

[orangetea]

I thought you can drop the X's because most of the reactants didn't shift to products?

yes my bad reactant side is bigger than product side which is why your Keq is so small. And yea you can just ignore the shift.