Scientific Tutor

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.

N_2(g) + 3 H_2(g) <—-> 2 NH_3(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?

 

COMPLETE ANSWER:[NH₃] = 1.5M

 

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.

H₂SO_4(aq) <—–> H_2(aq) + S_(s) + 2 O_2(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?

 

COMPLETE ANSWER:[H₂ ] = 0.5M

 

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 H_2(g) + O_2(g)

Answer: [ H₂O] = 1M

 

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.

MgBr_2(aq) + 2 NaI_(aq) <——> MgI_2(aq) + 2 NaBr_(aq)

Answer: [ NaBr] = 4M

 

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.

C_4(s) + 4 O_2(g) <—-> 4 CO_2(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 N₂O_3(g) <——> 2 N_2(g) + 3 O_2(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 OF_2(g) + 2 NH_3(g) <—> N₂F_4(g) + O_2(g) + 3 H_2(g)

Answer: K = 13.5