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### ICE BOX and coefficients

Posted: **Fri Jan 25, 2019 12:12 pm**

by **Dakota_Campbell_1C**

When doing ice box problems what are you supposed to do with the coefficient in front of a molecule? do you add it to the change or do you multiply? Also when you are adding Mols to the equation do you change it for the base equation?

### Re: ICE BOX and coefficients

Posted: **Fri Jan 25, 2019 12:22 pm**

by **Josephine Lu 4L**

the coefficient in front of the molecule is multiplied to x when you fill out the change values in the ICE box. For example, if a molecule has the coefficient 2, it would change by +/- 2x.

### Re: ICE BOX and coefficients

Posted: **Fri Jan 25, 2019 12:22 pm**

by **Julia Go 2L**

Using the equation N2(g) + 3H2(g) <--> 2NH3(g) as an example. The coefficients in front of the reactant and product affects the change in concentration. If you started with 0.1M of N2 and H2, in the ICE box you would put the initial concentration as:

N2 = 0.1

H2 = 0.1

NH3 = 0

Next for the change in concentration, it would be:

N2 = -x

H2 = -3x

NH3 = +2x

Finally the equilibrium concentration is:

N2 = 0.1-x

H2 = 0.1-3x

NH3 = 2x

### Re: ICE BOX and coefficients

Posted: **Fri Jan 25, 2019 1:04 pm**

by **Selina Bellin 2B**

if you have a coefficient in ur reaction, in the ICE box, multiply ur equilibrium (for example, x) by the coefficient. also, don't forget in ur K expression to use that coefficient as its exponent

### Re: ICE BOX and coefficients

Posted: **Sun Jan 27, 2019 11:15 pm**

by **Michelle Song 1G**

When there are coefficients in front of the molecules you multiply it by x (the change).