Please help with 1 and 2 (physical chemistry) 1) Consider the reaction: C2H4(g)

Question

Please help with 1 and 2 (physical chemistry)

1) Consider the reaction: C2H4(g) + Cl2(g) <---> C2H4Cl2(g). For the following scenarios below, write "left" if the reaction will shift to the left, "right" if the reaction will shift to the right, or "no effect" if the reaction will not shift in either direction.

(i) adding Cl2

(ii) increasing the reaction volume Blank 2 (iii) adding an inert gas to the reaction mixture

(iv) removing C2H4Cl2 from the reaction

(v) lowering the reaction temperature

2)

Consider the reaction: 2 H2S(g) <----> 2 H2(g) + S2(g)     KP = 2.4 x 10-4 at 1073 K

A reaction mixture contains 0.112 atm of H2, 0.055 atm of S2, and 0.445 atm of H2S.

(i) Is the reaction mixture at equilibrium? (Enter yes or no)

(ii) If not, in what direction will the reaction proceed? (Enter left or right)

Explanation / Answer

1. i) C2H4(g) + Cl2(g) ---> C2H4Cl2(g) ; [ According to Le Châtelier's principle, addition of excess reactant shift the equilibrium to the right]

ii) C2H4(g) + Cl2(g) <--- C2H4Cl2(g) ; [ Partial pressure of the components decrease by increasing volume, the equilibrium of the above reaction will shift to the left, because the reactant side has greater number of moles than does the product side. The system tries to counteract the decrease in partial pressure of gas molecules by shifting to the side that exerts greater pressure.]

iii) An inert gas (or noble gas) such as helium is one that does not react with other elements or compounds. Adding an inert gas into a gas-phase equilibrium at constant volume does not result in a shift. Here the total pressure of the system increases, the total pressure does not have any effect on the equilibrium constant; rather, it is a change in partial pressures that will cause a shift in the equilibrium. If, however, inert gas is added at a constant pressure, i.e.,if the volume is allowed to increase in the process, the partial pressures of all gases would be decreased resulting in a shift towards the side with the greater number of moles of gas.

iv) C2H4(g) + Cl2(g) ---> C2H4Cl2(g).

v) In the reaction is exothermic, increase in temperature decreases the equilibrium constant, K (equilibrium shift to the left) whereas, in endothermic reactions, increase in temperature increases the K (shift to the right) value.

2. 2 H2S(g) <----> 2 H2(g) + S2(g)  

Kp= [S2][H2]2/[H2S]2 = 0.055 * 0.122

=3.48*10-3

i) No, as the Kp values do not match.

ii) 3.48*10-3>2.4 x 10-4 , so the reaction will proceed to the left to reduce the Kp value.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

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