Reaction BaO2 (s)\(\rightleftharpoons\) BaO (s) + O2(g); ∆H = + ve. In equilibrium condition. The pressure of O2 depends on :
Increase mass of BaO2
Increase mass of BaO2 and BaO both
The correct option is (C) : Increase temp. on Eqm.
The following reaction is given: BaO2 (s)\(\rightleftharpoons\)BaO (s)+ O2 (g); H= +ive H is positive in this reaction because. Its nature is endothermic as a result. By using Le Chatelier's principle, we can show that the reaction moves ahead as the temperature rises. As a result, as the temperature rises at equilibrium, the pressure of oxygen will also rise.
The equilibrium constant may be defined as the ratio between the product of the molar concentrations of the products to that of the product of the molar concentrations of the reactants with each concentration term raised to a power equal to the stoichiometric coefficient in the balanced chemical reaction.
The equilibrium constant at a given temperature is the ratio of the rate constant of forwarding and backward reactions.
Kequ = kf/kb = [C]c [D]d/[A]a [B]b = Kc
where Kc, indicates the equilibrium constant measured in moles per litre.
For reactions involving gases: The equilibrium constant formula, in terms of partial pressure will be:
Kequ = kf/kb = [[pC]c [pD]d]/[[pA]a [pB]b] = Kp
Where Kp indicates the equilibrium constant formula in terms of partial pressures.
Medium Kc/Kp values indicate optimum product formation.
The equilibrium constant is the ratio of the concentrations raised to the stoichiometric coefficients. Therefore, the unit of the equilibrium constant = [Mole L-1]△n.
where, ∆n = sum of stoichiometric coefficients of products – a sum of stoichiometric coefficients of reactants.