Calculating Kc from Known Equilibrium Amounts
Example:Â Calculate the value of the equilibrium constant, K_{c}Â , for the system shown, if 0.1908 moles of CO_{2}, 0.0908 moles of H_{2}, 0.0092 moles of CO, and 0.0092 moles of H_{2}O vapor were present in a 2.00 L reaction vessel were present at equilibrium.
Calculating K from Initial Amounts and One Known Equilibrium Amount
Example:Â Â Initially, a mixture of 0.100 M NO, 0.050 M H_{2}, 0.100 M H_{2}Â O was allowed to reach equilibrium (initially there was no N_{2}Â ).Â At equilibrium the concentration of NO was found to be 0.062 M.Â Determine the value of the equilibrium constant, K_{c}Â , for the reaction:
NO | H_{2} | N_{2} | H_{2}O | |
Initial Concentration (M) | 0.100 | 0.0500 | 0 | 0.100 |
Change in Concentration (M) |
– 2 x |
– 2 x |
+ x |
+ 2 x |
Equilibrium Concentration (M) | 0.062 |
The change in concentration of the NO was (0.062 M – 0.100M) = – 0.038 M.Â Thus -2 x = – 0.038 and x = 0.019. Note: the negative sign indicates a decreasing concentration, not a negative concentration.Â The changes in the other species must agree with the stoichiometry dictated by the balance equation.Â The hydrogen will also change by – 0.038 M, while the nitrogen will increase by + 0.019 M and the water will increase by + 0.038 M.Â From these changes we can complete the chart to find the equilibrium concentrations for each species.
NO | H_{2} | N_{2} | H_{2}O | |
Initial Concentration (M) | 0.100 | 0.0500 | 0 | 0.100 |
Change in Concentration (M) | – 0.038 | – 0.038 | + 0.019 | + 0.038 |
Equilibrium Concentration (M) | 0.062 | 0.012 | 0.019 | 0.138 |
Calculating Kc from Known Initial Amounts and the Known Change in Amount of One of the Species
Example:Â A flask is charged with 3.00 atm of dinitrogen tetroxide gas and 2.00 atm of nitrogen dioxide gas at 25^{o}C and allowed to reach equilibrium.Â It was found that the pressure of the nitrogen dioxide decreased by 0.952 atm.Â Estimate the value of K_{p}Â for this system:
N_{2}O_{4} | NO_{2} | |
Initial Pressure (atm)(kpa) | 3.00 | 2.00 |
Change in Pressure (atm)(kpa) | + 0.476 | – 0.952 |
Equilibrium Pressure (atm)(kpa) | 3.476 | 1.048 |
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