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 N_{2} was 2M, and the initial concentration of H_{2} was 7M. What would be the concentration of NH_{3} be at equilibrium if the concentration of N_{2} 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.
N_{2(g)} + 3 H_{2(g)} <— | —> 2 NH_{3(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 NH_{3} has a concentration of 0M. The problem also gives us the equilibrium concentration of N_{2}.
N_{2(g)} + 3 H_{2(g)} <— | —> 2 NH_{3(g)} | |
INITIAL | 2M 7M | 0M |
CHANGE | ||
EQUILIBRIUM | 1.5M |
Now look at the difference between the initial concentration and the equilibrium concentration of N_{2}. The difference is -0.5M so we fill that in on the ICE table in the change section.
N_{2(g)} + 3 H_{2(g)} <— | —> 2 NH_{3(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 N_{2} was -0.5M, the change of H_{2} must be 3 * -0.5M.
CHANGE of H_{2} = 3 * -0.5M = -1.5M
The only difference with NH_{3} is the sign. The sign of NH_{3} is positive because the ICE table side that starts at 0M always gains some.
CHANGE of NH3 = 2 * +0.5M = 1M
N_{2(g)} + 3 H_{2(g)} <— | —> 2 NH_{3(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.
N_{2(g)} + 3 H_{2(g)} <— | —> 2 NH_{3(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 NH_{3} at equilibrium?
COMPLETE ANSWER:[NH_{3}] = 1.5M
VIDEO Solving ICE Tables Demonstrated Example 2: If the initial concentration of H_{2}SO_{4} was 4M. What would be the concentration of H_{2 }be at equilibrium if the concentration of O_{2 } at equilibrium is 1M? Use the balanced chemical equation below.
H_{2}SO_{4(aq)} <—–> H_{2(aq)} + S_{(s)} + 2 O_{2(aq)}
First set up the ice table.
H_{2}SO_{4(aq) }<– | –> H_{2(aq)} + S_{(s)} + 2 O_{2(aq)} | |
INITIAL | ||
CHANGE | ||
EQUILIBRIUM |
Cross out the S because it is a solid and it will not take part in any of our calculations.
H_{2}SO_{4(aq) }<– | –> H_{2(aq)} + S_{(s)} + 2 O_{2(aq)} | |
INITIAL | ||
CHANGE | ||
EQUILIBRIUM |
Then start to fill in the information from the problem.
H_{2}SO_{4(aq) }<– | –> H_{2(aq)} + S_{(s)} + 2 O_{2(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 O_{2}.
H_{2}SO_{4(aq) }<– | –> H_{2(aq)} + S_{(s)} + 2 O_{2(aq)} | |
INITIAL | 4M | 0M 0M |
CHANGE | +1M | |
EQUILIBRIUM | 1M |
How does the change in the O_{2} relate the to change in other chemicals? Since O_{2} has a 2 coefficient then it has twice the change of H_{2} and H_{2}SO_{4}. Therefore…
+1M / 2 = + 0.5M = H_{2}
-1M / 2 = -0.5M = H_{2}SO_{4}
H_{2}SO_{4(aq) }<– | –> H_{2(aq)} + S_{(s)} + 2 O_{2(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.
H_{2}SO_{4(aq) }<– | –> H_{2(aq)} + S_{(s)} + 2 O_{2(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 H_{2 }at equilibrium?
COMPLETE ANSWER:[H_{2 }] = 0.5M
PRACTICE PROBLEMS: Solve the ICE table problems below.
If the initial concentration of O_{2 } was 1.5M, and the initial concentration of H_{2} was 2M. What would be the concentration of H_{2}O be at equilibrium if the concentration of O_{2} at equilibrium is 1M? Use the balanced chemical equation below.
2 H_{2}O_{(g)} <—-> 2 H_{2(g)} + O_{2(g)}
Answer: [ H_{2}O] = 1M
If the initial concentration of MgBr_{2 }was 5M, and the initial concentration of NaI was 8M. What would be the concentration of NaBr be at equilibrium if the concentration of MgBr_{2 } 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 CO_{2 }was 10M. What would be the concentration of CO_{2 } be at equilibrium if the concentration of O_{2 }at equilibrium is 4M? Use the balanced chemical equation below.
C_{4(s)} + 4 O_{2(g)} <—-> 4 CO_{2(g)}
Answer: [ CO_{2 }] = 6M
If the initial concentration of N_{2}O_{3 }was 0.4M. What would be the concentration of O_{2 }be at equilibrium if the concentration of N_{2}O_{3 }at equilibrium is 0.2M? Use the balanced chemical equation below.
2 N_{2}O_{3(g)} <——> 2 N_{2(g)} + 3 O_{2(g)}
Answer: [ O_{2 }] = 0.3M
If the initial concentration of OF_{2 }was 6M, and the initial concentration of NH_{3 }was 8M. What would be the equilibrium constant (K) if the concentration of H_{2 }at equilibrium is 6M? Use the balanced chemical equation below.
2 OF_{2(g)} + 2 NH_{3(g)} <—> N_{2}F_{4(g)} + O_{2(g)} + 3 H_{2(g)}
Answer: K = 13.5