What assumptions can you make with ICE or RICE tables?
When you have specific K values for equilibrium equations and ice tables most teachers and tests will allow you to make mathematically assumptions about them. The critical K values to look for are K values that are lower than 10^{-3} or higher than 10^{3}. This allows you to ignore the change values on one side of the ICE table when you are accounting for them in an equilibrium equation. You can do this because the values of change in those concentrations will be so small that it will not make a significant mathematical difference to the overall problem.
VIDEO Solving ICE Tables Demonstrated Example 3: If the initial concentration of H_{2}O was 3M, and the equilibrium constant was 2.3 * 10^{-3}. What would be the concentration of H_{2 } be at equilibrium? Use the balanced chemical equation below.
2 H_{2}O_{(g)} <—-> 2 H_{2(g)} + O_{2(g)}
First set up the ICE table and fill in the information from the problem.
2 H_{2}O_{(g)} <— | —> 2 H_{2(g)} + O_{2(g}_{)} | |
INITIAL | 3M | 0M 0M |
CHANGE | ||
EQUILIBRIUM |
Next, make an X value representing the change in concentration when it shifts to the equilibrium. Here is where one assumption comes in. It shifts to the right because the concentration values are zero on the right hand side.
2 H_{2}O_{(g)} <— | —> 2 H_{2(g)} + O_{2(g}_{)} | |
INITIAL | 10M | 0M 0M |
CHANGE | -2x | +2x +x |
EQUILIBRIUM |
Then add the initial and change together to get the equilibrium concentration.
2 H_{2}O_{(g)} <— | —> 2 H_{2(g)} + O_{2(g}_{)} | |
INITIAL | 10M | 0M 0M |
CHANGE | -2x | +2x +x |
EQUILIBRIUM | 3M-2x | 2x x |
Now construct the equilibrium equation from the equilibrium values you have.
K = | [ H_{2} ]^{2} [ O_{2} ] = | [ 2x ]^{2} [ x ] = | 2.3 * 10^{-3} |
[ H_{2}O ]^{2} | [ 3M-2x ]^{2} |
Here is where another assumption comes in. Since our K is 10^{-3} we can assume that the -2x from the bottom of the equation does not matter and therefore we can remove it from the equation.
[ 2x ]^{2} [ x ] = | 2.3 * 10^{-3} |
[ 3M-2x ]^{2} |
Remove it.
[ 2x ]^{2} [ x ] = | 2.3 * 10^{-3} |
[ 3M ]^{2} |
The problem now becomes much easier to solve for. Our goal now is to solve for x. Start by applying the exponents.
[ 4x^{2} ] [ x ] = | 2.3 * 10^{-3} |
[ 9 ] |
Multiply the top.
[ 4x^{3} ] = | 2.3 * 10^{-3} |
[ 9 ] |
Multiply both sides by 9.
9 * [ 4x^{3} ] = | 2.3 * 10^{-3} * 9 |
[ 9 ] |
Cross out the 9 on the left side and multiply the values on the right side.
9 * [ 4x^{3} ] = | 0.0207 |
[ 9 ] |
Simplify
[ 4x^{3} ] = | 0.0207 |
1 |
Divide both sides by 4.
[ 4x^{3} ] = | 0.0207 |
4 | 4 |
Cross out the 4 on the left side and divide the 4 on the right side.
[ 4x^{3} ] = | 0.005175 |
4 |
Simplify
[ x^{3} ] = | 0.005175 |
1 |
Take the cube (3) root of both sides.
[ x ] = | 0.176 |
1 |
Now that we found x we can solve for what H_{2} will be at equilibrium, which is 2x.
[ H_{2} ] = 0.176 * 2 = 0.352
COMPLETE ANSWER:[ H_{2 }] = 0.352 M
You can check this answer by putting the concentrations back into the original equation and then finding out what K value you get with them. The difference in the K value you come up with is about 3 times but at the levels of 10^{-3} this is pretty insignificant.
VIDEO Solving ICE Tables Demonstrated Example 4: If the initial concentration of HCl was 0.5M, and the equilibrium constant was 6.3 * 10^{4}. What would be the concentration of Cl_{2} be at equilibrium? Use the balanced chemical equation below.
Cl_{2(g)} + H_{2(g)} <—> 2 HCl_{(g)}
First set up the ICE table and fill in the information from the problem.
Cl_{2(g)} + H_{2(g)} <— | —> 2 HCl_{(g)} | |
INITIAL | 0M 0M | 0.5M |
CHANGE | ||
EQUILIBRIUM |
Next, make an X value representing the change in concentration when it shifts to the equilibrium. Here is where one assumption comes in. It shifts to the left because the concentration values are zero on the right hand side.
Cl_{2(g)} + H_{2(g)} <— | —> 2 HCl_{(g)} | |
INITIAL | 0M 0M | 0.5M |
CHANGE | +x +x | -2x |
EQUILIBRIUM |
Then add the initial and change together to get the equilibrium concentration.
Cl_{2(g)} + H_{2(g)} <— | —> 2 HCl_{(g)} | |
INITIAL | 0M 0M | 0.5M |
CHANGE | +x +x | -2x |
EQUILIBRIUM | x x | 0.5M-2x |
Now construct the equilibrium equation from the equilibrium values you have.
K = | [ HCl ]^{2} = | [ 0.5M-2x]^{2} = | 6.3 * 10^{4} |
[ Cl_{2} ] [ H_{2 }] | [ x ] [ x ] |
Here is where another assumption comes in. Since our K is 10^{4} we can assume that the -2x from the top of the equation does not matter and therefore we can remove it from the equation.
[ 0.5M-2x]^{2} = | 6.3 * 10^{4} |
[ x ] [ x ] |
Remove it.
[ 0.5M]^{2} = | 6.3 * 10^{4} |
[ x ] [ x ] |
The problem now becomes much easier to solve for. Our goal now is to solve for x. Start by applying the exponents.
[ 0.25 ] = | 6.3 * 10^{4} |
[ x ] [ x ] |
Multiply the bottom left.
[ 0.25 ] = | 6.3 * 10^{4} |
x^{2} |
Multiply both sides by x^{2}.
x^{2} * [ 0.25 ] = | 6.3 * 10^{4} * x^{2} |
x^{2} |
Cross out the x^{2} on the left side.
x^{2} * [ 0.25 ] = | 6.3 * 10^{4} * x^{2} |
x^{2} |
Simplify
[ 0.25 ] = | 6.3 * 10^{4} * x^{2} |
1 |
Divide both sides by 6.3 * 10^{4}.
[ 0.25 ] = | 6.3 * 10^{4} * x^{2} |
6.3 * 10^{4} | 6.3 * 10^{4} |
Cross out 6.3 * 10^{4} on the right side.
[ 0.25 ] = | 6.3 * 10^{4} * x^{2} |
6.3 * 10^{4} | 6.3 * 10^{4} |
Simplify
[ 0.25 ] = | x^{2} |
6.3 * 10^{4} |
Divide the left side.
3.97 * 10^{-6} = | x^{2} |
1 |
Take the square (2) root of both sides.
1.99 * 10^{-3} = | x |
1 |
COMPLETE ANSWER:[ Cl_{2 }] = 1.99 * 10^{-3} M
PRACTICE PROBLEMS: Solve the equilibrium ICE table problems below.
If the initial concentration of HBr was 1.5M, and the equilibrium constant was 3.2 * 10^{-4}. What would be the concentration of H_{2 } be at equilibrium? Use the balanced chemical equation below.
2 HBr_{(g)} <—> H_{2(g)} + Br_{2(g)}
Answer: 0.0268M
If the initial concentration of BaCl_{2} was 2M, and the equilibrium constant was 4.6 * 10^{5}. What would be the concentration of HCl be at equilibrium? Use the balanced chemical equation below.
Ba(OH)_{2(aq)} + 2 HCl_{(aq)} <——> 2 H_{2}O_{(l)} + BaCl_{2(aq)}
Answer: 0.0206M (it is 2 * the x number)
If the initial concentration of HBr was 0.035M, and the equilibrium constant was 8.7 * 10^{-8}. What would be the concentration of O_{2 }be at equilibrium? Use the balanced chemical equation below.
H_{2}SO_{4(aq)} <—–> H_{2(aq)} + S_{(s)} + 2 O_{2(aq)}
Answer: 0.00196M (it is 2 * the x number)