12. Consider the following equilibrium process: Predict the direction of the shi

Question

12. Consider the following equilibrium process: Predict the direction of the shift in equilibrium when (a) the temperature is raised mixture with no volume change (b) more chlorine gas is added to the reaction no volume change (c) some PClyis removed from the mixture with the pressure on the gases is increased by decreasing the volume (e) a catalyst is added to the reaction mixture 13. Consider the reaction AH 1982 kJ/mol Comment on the changes in the concentrations of so2.o2, and Sos at equilibrium if we (a) increase the temperature n increase incr ars, SO2 .decrease (b) increase the pressure by decreasing the volume So3 Or deciosc ncrease SO2 at a constant volume (c) Soz incre se, a decrease Sor ncrease (d) add a catalyst (e) add helium at a constant volume. 14. In the uncatalyzed reaction the pressure gases at equilibrium are PNo 0.377 and PNo, 156 atm at 100°C What would happen these pressures if a catalyst were added to the mixture?

Explanation / Answer

Answer – 12) We are given,

PCl5(g) <----> PCl3(g) + Cl2(g)    Ho = 92.5 kJ/mol

We know as per Le chatelier principle, as the concentration of reactant species increase then there is equilibrium shifted towards product side means right side and there is formed more product.

a) The given reaction absorbed heat, so it is endothermic reaction. As we raised the temperature then there is more heat absorbed and there is more favorable for forward reaction. So there is more product formed, means equilibrium will shift towards product means right side.

b) As per Le chatelier principle we know as we increase the species in the one side of the reaction then the equilibrium shifted towards other side. When we added the Cl2(g) more then there is reverse reaction and formed more reactant, so equilibrium will shift towards reactant means left side

c) As per Le chatelier principle we know as we removing the species from the respective side of the reaction then the equilibrium shifted towards same side. When we remove the PCl5(g) then there is formed more reactant, so equilibrium will shift towards reactant means left side

d) As per Le chatelier principle we know as we increase the pressure of gases reaction then the equilibrium shifted towards the fewer moles. In this reaction there are fewer mole in the reactant side, so equilibrium will shift towards reactant means left side.

e) We know when there is reaction has no used any catalyst then upon the addition of catalyst there is no any effect on the equilibrium. In this reaction there no any catalyst used, so equilibrium not shifted left as well as right and it is remaining in between both

13) We are given,

2SO2(g) + O2(g)   <----> 2SO3(g)   Ho = -198.2 kJ/mol

We describe the all condition in the above question which are affected on the equilibrium. So following are the effect affected on the equilibrium concentration

a) Increase the temp – The given reaction is exothermic reaction, so as we increase the temp reaction gets reversed and equilibrium will shift towards reactant means left side.

At equilibrium

[SO2(g)] = increase

[O2(g)] = increase

[SO3(g)] = decrease

b) Increase the pressure – The fewer mole is on the product side, so equilibrium will shift towards product means right side.

At equilibrium

[SO2(g)] = decrease

[O2(g)] = decrease

[SO3(g)] = increase

c) Increase the SO2 – As we increase the SO2 concentration then equilibrium will shift towards product means right side –

At equilibrium

[SO2(g)] = decrease

[O2(g)] = decrease

[SO3(g)] = increase

d) Add catalyst – When the addition of catalyst there is no any effect on the equilibrium.

At equilibrium

[SO2(g)] = No change

[O2(g)] = No change

[SO3(g)] = No change

e) Add helium at constant volume – In the reaction there is no any Helium atom and not it is used as catalyst, so there is no any effect on the equilibrium.

At equilibrium

[SO2(g)] = No change

[O2(g)] = No change

[SO3(g)] = No change

14) We are given unanalyzed reaction -

N2O4(g) <----> 2NO2(g)

We are given unanalyzed reaction and there is equilibrium pressure of NO2 is higher than N2O4(g), so when we added catalyst then there is reaction rate increase and the reaction goes reverse, means from NO2(g) to N2O4(g). The equilibrium pressure of N2O4(g) increase where for NO2(g) it is decrease.

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.