## Chem – College: Simple Ice Tables

What are ICE or RICE tables and how do you use them in equilibrium?

EXPLATION

VIDEO Solving ICE Tables Demonstrated Example 1: If the initial concentration of N2 was 2M, and the initial concentration of H2 was 7M. What would be the concentration of NH3 be at equilibrium if the concentration of N2 at equilibrium is 1.5M? Use the balanced chemical equation below.

N2(g) + 3 H2(g) <—-> 2 NH3(g)

First set up the ice table.

 N2(g)  +   3 H2(g) <— —> 2 NH3(g) INITIAL CHANGE EQUILIBRIUM

Then start to fill in the information from the problem. It gives us the initial values of the molecules so we write those directly below the molecules in the equation. If it does not say the initial concentration of a particular molecule, then that means that the molecule has an initial concentration of 0M. In this case the NH3 has a concentration of 0M. The problem also gives us the equilibrium concentration of N2.

 N2(g)   +   3 H2(g) <— —> 2 NH3(g) INITIAL 2M            7M 0M CHANGE EQUILIBRIUM 1.5M

Now look at the difference between the initial concentration and the equilibrium concentration of N2. The difference is -0.5M so we fill that in on the ICE table in the change section.

 N2(g)   +    3 H2(g) <— —> 2 NH3(g) INITIAL 2M             7M 0M CHANGE -0.5M EQUILIBRIUM 1.5M

How does that relate to the rest of the ICE table? Once you fill in the change you can look back to the coefficients in the problem. The change in each concentration will be in proportion to the coefficients of the problem.

Therefore, if the change of N2 was -0.5M, the change of H2 must be 3 * -0.5M.

CHANGE of H2 = 3 * -0.5M = -1.5M

The only difference with NH3 is the sign. The sign of NH3 is positive because the ICE table side that starts at 0M always gains some.

CHANGE of NH3 = 2 * +0.5M = 1M

 N2(g)  +  3 H2(g) <— —> 2 NH3(g) INITIAL 2M          7M 0M CHANGE -0.5M     -1.5M +1.5M EQUILIBRIUM 1.5M

If we add together the INITIAL and CHANGE categories on the ICE table we then get the EQUILIBRIUM values for all of the molecules.

 N2(g)    +   3 H2(g) <— —> 2 NH3(g) INITIAL 2M             7M 0M CHANGE -0.5M        -1.5M +1.5M EQUILIBRIUM 1.5M          5.5M 1.5M

Therefore, we have answered our question: What is the concentration of NH3 at equilibrium?

VIDEO Solving ICE Tables Demonstrated Example 2: If the initial concentration of H2SO4 was 4M. What would be the concentration of H2 be at equilibrium if the concentration of O2 at equilibrium is 1M? Use the balanced chemical equation below.

H2SO4(aq) <—–> H2(aq) + S(s) + 2 O2(aq)

First set up the ice table.

 H2SO4(aq) <– –> H2(aq) + S(s) + 2 O2(aq) INITIAL CHANGE EQUILIBRIUM

Cross out the S because it is a solid and it will not take part in any of our calculations.

 H2SO4(aq) <– –> H2(aq) + S(s) + 2 O2(aq) INITIAL CHANGE EQUILIBRIUM

Then start to fill in the information from the problem.

 H2SO4(aq) <– –> H2(aq) + S(s) + 2 O2(aq) INITIAL 4M 0M                  0M CHANGE EQUILIBRIUM 1M

Now look at the difference between the initial concentration and the equilibrium concentrations. Fill in the change for the O2.

 H2SO4(aq) <– –> H2(aq) + S(s) + 2 O2(aq) INITIAL 4M 0M                   0M CHANGE +1M EQUILIBRIUM 1M

How does the change in the O2 relate the to change in other chemicals? Since O2 has a 2 coefficient then it has twice the change of H2 and H2SO4. Therefore…

+1M / 2 = + 0.5M = H2

-1M / 2 = -0.5M = H2SO4

 H2SO4(aq) <– –> H2(aq) + S(s) + 2 O2(aq) INITIAL 4M 0M                   0M CHANGE -0.5M +0.5M              +1M EQUILIBRIUM 1M

If we add together the INITIAL and CHANGE categories on the ICE table we then get the EQUILIBRIUM values for all of the molecules.

 H2SO4(aq) <– –> H2(aq) + S(s) + 2 O2(aq) INITIAL 4M 0M                   0M CHANGE -0.5M +0.5M               +1M EQUILIBRIUM 3.5M 0.5M                 1M

Therefore, we have answered our question: What is the concentration of H2 at equilibrium?

PRACTICE PROBLEMS: Solve the ICE table problems below.

If the initial concentration of O2 was 1.5M, and the initial concentration of H2 was 2M. What would be the concentration of H2O be at equilibrium if the concentration of O2 at equilibrium is 1M? Use the balanced chemical equation below.

2 H2O(g) <—-> 2 H2(g) + O2(g)

If the initial concentration of MgBr2 was 5M, and the initial concentration of NaI was 8M. What would be the concentration of NaBr be at equilibrium if the concentration of MgBr2 at equilibrium is 3M? Use the balanced chemical equation below.

MgBr2(aq) + 2 NaI(aq) <——> MgI2(aq) + 2 NaBr(aq)

If the initial concentration of CO2 was 10M. What would be the concentration of CO2 be at equilibrium if the concentration of O2 at equilibrium is 4M? Use the balanced chemical equation below.

C4(s) + 4 O2(g) <—-> 4 CO2(g)

Answer: [ CO2 ] = 6M

If the initial concentration of N2O3 was 0.4M. What would be the concentration of O2 be at equilibrium if the concentration of N2O3 at equilibrium is 0.2M? Use the balanced chemical equation below.

2 N2O3(g) <——> 2 N2(g) + 3 O2(g)

Answer: [ O2 ] = 0.3M

If the initial concentration of OF2 was 6M, and the initial concentration of NH3 was 8M. What would be the equilibrium constant (K) if the concentration of H2 at equilibrium is 6M? Use the balanced chemical equation below.

2 OF2(g) + 2 NH3(g) <—> N2F4(g) + O2(g) + 3 H2(g)