In the first question the reaction is at equilibrium. When the temperature is decreased, more N2O4 if formed and K increases. When the temperature is increased, more NO2 is formed and K decreases.
Removing O₂, means removing one of the reactants and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.
Explanation:
The principle that explains how changes in temperature, Concentration and Pressure of reactants or products of a reaction at equilibrium affect the equilibrium position of the reaction is the Le Chatelier's principle.
The Principle explains that a system/process if a system/process which is at equilibrium is disturbed/perturbed/constrained by one or more changes (in concentration, pressure or temperature), the system would shift the equilibrium position to counteract the effects of this change.
Removing O₂, means removing one of the reactants (changing its concentration) and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.
As with the hydrogen-ion concentration, the concentration of the hydroxide ion can be expressed logarithmically by the pOH. The pOH of a solution is the negative logarithm of the hydroxide-ion concentration. pOH=−log[OH−] The pH of a solution can be related to the pOH.
