Physical Sciences Grade 10-11-12.
Calculating Kc from Known Equilibrium Amounts
Example: Calculate the value of the equilibrium constant, Kc , for the system shown, if 0.1908 moles of CO2, 0.0908 moles of H2, 0.0092 moles of CO, and 0.0092 moles of H2O 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: Initially, a mixture of 0.100 M NO, 0.050 M H2, 0.100 M H2 O was allowed to reach equilibrium (initially there was no N2 ). At equilibrium the concentration of NO was found to be 0.062 M. Determine the value of the equilibrium constant, Kc , for the reaction:
NO | H2 | N2 | H2O | |
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. Thus -2 x = – 0.038 and x = 0.019. Note: the negative sign indicates a decreasing concentration, not a negative concentration. The changes in the other species must agree with the stoichiometry dictated by the balance equation. 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. From these changes we can complete the chart to find the equilibrium concentrations for each species.
NO | H2 | N2 | H2O | |
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: A flask is charged with 3.00 atm of dinitrogen tetroxide gas and 2.00 atm of nitrogen dioxide gas at 25oC and allowed to reach equilibrium. It was found that the pressure of the nitrogen dioxide decreased by 0.952 atm. Estimate the value of Kp for this system:
N2O4 | NO2 | |
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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