MedVision ad

On ICE table (1 Viewer)

Sp3ctre

Active Member
Joined
Nov 29, 2016
Messages
187
Location
NSW
Gender
Male
HSC
2017
They're used for finding equilibrium constants (K) when you don't have sufficient information to calculate K directly using the formula.
You'll usually need to use it when the question gives you a chemical equation and some small bits of information such as initial moles or concentration, but without giving you the equilibrium concentrations of everything in the equation (which is what you need when using the formula).

This is a neat example I yoinked from another forum which could help you understand it a bit better:

0.10 mol of N2O4 is allowed to come to equib in a 2.0 L flask according to the following reaction:
N2O4(g) <---> 2NO2(g)
At equib, there are 0.060 mol of NO2. Determine K


To determine K, we need the equilibrium concentrations of both the reactants and products (I'm assuming you're familiar with the K formula already)
However, all we are given from this question is the initial moles of N2O4 and the final (equilibrium) moles of NO2. This is when we need to set up the ICE table.

N2O4NO2
I - initial n0.10
C - change in n
E - equilibrium n0.060

The most obvious place to start is the initial moles of 2NO2. There shouldn't be products initially, so initial n(NO2) = 0.

Now we can work out the change in moles of 2NO2 since we have both initial n(NO2) and equilibrium n(NO2).
Change in n(2NO2) = 0.060 - 0 = +0.060 mol. Pretty straightforward!

Now pay attention to the mole ratio of the equation. It's a 1:2 ratio, which means for every mole of N2O4, 2 moles of NO2 are produced.
Therefore, change in n(N2O4) = 0.060 / 2 = -0.030 mol. Keep in mind that this value is negative since N2O4 is the reactant.

Finally, we can now easily find the moles of N2O4 at equilibrium.
n(N2O4) at equilibrium = 0.10 - 0.030 = 0.07 mol.


N2O4NO2
I - initial in n0.100
C - change in n-0.030+0.060
E - equilibrium n0.070.060

That's all you need to use the ICE table for. But to finish the question, you need to use the equilibrium moles that we found using the ICE table to find the equilibrium concentrations using n = c / v, then use those concentrations to find K. Hope that helps!
 
Last edited:

NexusRich

Member
Joined
Jul 6, 2019
Messages
87
Gender
Male
HSC
2021
They're used for finding equilibrium constants (K) when you don't have sufficient information to calculate K directly using the formula.
You'll usually need to use it when the question gives you a chemical equation and some small bits of information such as initial moles or concentration, but without giving you the equilibrium concentrations of everything in the equation (which is what you need when using the formula).

This is a neat example I yoinked from another forum which could help you understand it a bit better:

0.10 mol of N2O4 is allowed to come to equib in a 2.0 L flask according to the following reaction:
N2O4(g) <---> 2NO2(g)
At equib, there are 0.060 mol of NO2. Determine K


To determine K, we need the equilibrium concentrations of both the reactants and products (I'm assuming you're familiar with the K formula already)
However, all we are given from this question is the initial moles of N2O4 and the final (equilibrium) moles of NO2. This is when we need to set up the ICE table.

N2O42NO2
I - initial n0.10
C - change in n
E - equilibrium n0.060
The most obvious place to start is the initial moles of 2NO2. There shouldn't be products initially, so initial n(2NO2) = 0.


Now we can work out the change in moles of 2NO2 since we have both initial n(2NO2) and equilibrium n(2NO2).
Change in n(2NO2) = 0.060 - 0 = +0.060 mol. Pretty straightforward!

Now pay attention to the mole ratio of the equation. It's a 1:2 ratio, which means for every mole of N2O4, 2 moles of NO2 are produced.
Therefore, change in n(N2O4) = 0.060 / 2 = -0.030 mol. Keep in mind that this value is negative since N2O4 is the reactant.

Finally, we can now easily find the moles of N2O4 at equilibrium.
n(N2O4) at equilibrium = 0.10 - 0.030 = 0.07 mol.


N2O42NO2
I - initial in n0.100
C - change in n-0.030+0.060
E - equilibrium n0.070.060
That's all you need to use the ICE table for. But to finish the question, you need to use the equilibrium moles that we found using the ICE table to find the equilibrium concentrations using n = c / v, then use those concentrations to find K. Hope that helps!
Thank you so much!!!!! I understand it now :cool:
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top