# How to use ICE charts to find equilibrium concentration

## Step method to

Q1 If 1.00 mol each of carbon dioxide and hydrogen is initially injected into a 10.0-L reaction chamber at 986°C, what would be the concentrations of each entity at equilibrium?

CO_{2(g)} +H_{2(g)}→CO_{(g)} +H_{2}O_{(g)} *K *= 1.60 for 986°C

( temperature is just to confuse you. Ignore it)

Step1:Convert the concentration into mol/l( if the concentration is already given in mol/l skip this step)

To get mol/l divide the number of moles with total volume given.1 mol/10 l=0.1 mol/l

Step2: Construct ICE chart by writing the equation.

Concentration(mol/l) |
CO2(g) +H2(g) →CO(g) +H2O(g) |

Initial |
0.1 0.1 0 0 |

Change |
-x* -x +x +x |

Equilibrium |
(0.1-x) (0.1-x) x x |

*Reactant's concentration goes down and product concentration goes up as the reaction proceeds.So for change in concentration we need to use -x for reactants and +x for products.

Step3: Write the equilibrium expression and solve it.

K=(CO)(H2O)/(CO2)(H2)

K=x×x/(0.1-x)(0.1-x)

K=x2/(0.1-x)2

1.60=x2/(0.1-x)2

Step 4:Square toot on both sides.

±1.26=x/(0.1-x)

1.26×0.1-1.26x=x

0.126=2.26x

x=0.126/2.26

x=0.55

(CO)=(H2O)=0.055 mol/l

(CO2)=(H2)=0.1-0.055=0.044 mol/l

Hint: the answer should be less than the initial concentration.If you get x value greater than the initial concentration then use negative value for K after square rooting it.

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