Chapter 1: Chemical Equilibrium Part 9 SABIS Grade 11 (Level M) Chemistry

1.6 Predicting New Equilibrium Concentrations

1.6.1 Le Chatelier’s Principle

We are not satisfied with only predicting the effect of a chemical or physical change on the equilibrium concentrations. We would also like to predict the direction of the effect (does it favor products or reactants?) and the magnitude of the effect (how much does it favor products or reactants?). The first desire, to know the qualitative effects, is answered by a generalization first proposed by a French chemist, Henry Louis Le Chatelier, and now called Le Chatelier’s Principle.

Le Chatelier sought regularities among a large amount of experimental data concerning equilibria.
To summarize the regularities he found, he made this generalization: If an equilibrium system is subjected to a change, processes occur that tend to counteract partially the imposed change, if possible. This generalization has been found to be applicable to such a large number of systems that
it is now called a principle. Let us see how it applies to our examples.

Question

Fill in the blank.

Le

 principle states that: "if a system found at equilibrium, and is subjected to a change, processes tend
 to counteract partially the imposed change, if possible".

answer :

Chatelier 1.6.2 Concentration And Le Chatelier’s Principle Remember that we agreed to use the symbol [ ] to represent the concentration in moles  per liter. The formula within the brackets denotes the species. Thus the notation [X] stands for the equilibrium concentration X. If a soluble thiocyanate salt is added to an equilibrium solution containing both Fe3+(aq)  and SCN–(aq), the color of the complex ion increases: Fe3+(aq) + SCN–(aq) ⇌ FeSCN2+(aq) • Imposed change: [SCN–] (the concentration of SCN–) has been increased • To partially counteract the imposed change: system must decrease [SCN–] • This happens by Fe3+(aq) reacting with SCN–(aq) • Final concentration at equilibrium: [Fe3+(aq)] is lower, [SCN–] is lower (but remains  higher than the original), [FeSCN2+(aq)] is higher. The equilibrium shifts to the right. The change imposed on the system was an increase in the concentration of SCN–. This change can be counteracted in part by some Fe3+ and SCN– ions reacting to form  more FeSCN2+. Thus equilibrium shifts to the right. The same argument applies to an addition of iron (III) ion from a soluble iron (III) salt.  In each case, the formation of FeSCN2+ uses up a portion of the added reactant, partially counteracting the change. The equilibrium shifts to the right. What would happen to the [Fe3+(aq)] and [FeSCN2+(aq)] respectively if we add to the equilibrium  mixture few crystals of KSCN? The [Fe3+(aq)]              Choose...             increases.             decreases.          The [FeSCN2+(aq)]              Choose...             increases.             decreases.          answer : The [Fe3+(aq)] = decreases. The [FeSCN2+(aq)] = increases. Use Le Chatelier's principle to predict the changes in equilibrium concentrations in the following reaction: Fe3+(aq)  +  SCN−(aq)   ⇌   FeSCN2+(aq) What would happen to the [SCN−(aq)] and [FeSCN2+(aq)] respectively if Fe3+(aq) is removed by  adding phosphate ions? The [SCN−(aq)]              Choose...             increases.             decreases.          The [FeSCN2+(aq)]              Choose...             increases.             decreases.          answer : The [SCN−(aq)] = increases. The [FeSCN2+(aq)] = decreases. 1.6.3 Pressure And Le Chatelier’s Principle Case I: The partial pressure of reactant/products is changed H2O(g) ⇌ H2(g) + 1⁄2O2(g)         ΔH = + 242 kJ Predict what would happen if more H2 is injected into a reaction chamber containing the three gases. • Imposed change: [H2] is increased • To partially counteract the imposed change: system decreases [H2] • This happens by: H2 reacting with O2 to form more H2O(g) • Final concentrations at equilibrium: [H2O] is higher, [O2] is lower, [H2] is lower (but higher than the initial concentration). Equilibrium shifts to the left. Case II: Altering the volume H2O(g) ⇌ H2(g) + 1⁄2O2(g) Instead of altering the concentration of one individual component in an equilibrium  system, we can alter the concentration of all gaseous components by changing  the pressure by decreasing the volume of the reaction chamber. Let us double the total pressure by halving the volume. The concentration of all reactants doubles. The system cannot increase the volume (which we control),  but it can decrease the total of partial pressures. For this to happen, the total number of moles present must be reduced (this would happen if 0.5 moles  of products unite to form 1 mole of reactant). Hence, we can predict that increasing  the concentration of all components by increasing the partial pressure will shift the  state of equilibrium in favor of the formation of gaseous water. The equilibrium  shifts to the left. This is in accord with experiment. • Imposed change: pressure is doubled • To partially counteract the imposed change: system decreases total pressure • This happens by: H2 and O2 forming water (reducing total number of moles). • Final concentrations at equilibrium: moles of H2O are more, moles of O2 are fewer,  moles of H2 are fewer. Equilibrium shifts to the left. Case III: Adding helium gas into the reacting chamber A change in total pressure does not always shift equilibrium. If He gas is injected  into the chamber, the total pressure will increase. However, experiment shows that  the equilibrium concentrations are not affected.We explain this by the fact that the  partial pressures of the reactants have not changed, the concentrations of reactants and products have not changed. Case IV: Equal number of moles of reactants and products If we halve the volume of the reaction chamber for this reaction: CO(g) + NO2(g) ⇌ CO2(g) + NO(g) the concentrations (and all partial pressures) are all increased. However, whether equilibrium shifts to the right or to the left, the total number of moles in the reacting  chamber will not change (two moles of reactants form two moles of products).  Since the system cannot alter the number of moles by shifting in either direction,  the equilibrium will not change. Experimentally we can find that no change takes place. The equilibrium state is not affected by a pressure change (by changing volume) for any equilibrium gas mixture where the number of reactant molecules is the same as the number of  product molecules in the balanced reaction. Does Le Chatelier's Principle predict a change of equilibrium concentrations for the following reactions if the gas mixture is compressed? Give full explanation, showing the imposed change and how the system reacts? 2H2(g) + O2(g) ⇌ 2H2O(g) Imposed change: total pressure  .answer answer : increases next part