Consider this reaction, carried out at constant temperature and volume. 2SO_2(g)
ID: 501918 • Letter: C
Question
Consider this reaction, carried out at constant temperature and volume. 2SO_2(g) + O_2(g) 2SO_3(g) What is the effect of removing some SO_3 from a system initially at equilibrium? (A) The concentration of SO_2 decreases more than the concentration of O_2. (B) The concentration of SO_2 increases more than the concentration of O_2. (C) The concentration of SO_2 and O_2 remain the same. (D) The concentration of SO_2 and O_2 decrease equally. At 600K, K_P = 2.40 times 10^-8 for the reaction F_2(g) 2 F(g) If a system initially contains 0.500 atm of F_2(g) and no F(g), what will be the pressure of F(g) at equilibrium? A. 1.10 times 10^-4 atm B. 5.61 times 10^-5 atm C. 4.06 times 10^-4 atm D. 2.12 times 10^-6 atm The equilibrium constant for the reaction 2 NOCl(g) 2 NO(g) + Cl_2(g) is equal to 0.51 at a certain temperature. What is the equilibrium constant at the same temperature for the reaction NO(g) + 1/2 Cl_2(g) NOCl(g)? A. 1.4 B. 2.0 C. 3.8 D. 0.71 Consider this reaction, carried out at constant temperature and volume. PCl_5(g) PCl_3(g) + Cl_2(g) How can the position of equilibrium for this reaction be shifted to the right? (A) addition of a catalyst (B) removal of Cl_2 C) addition of an inert gas at constant volume (D) removal of PCl_5Explanation / Answer
24.[SO3] decreases:
*Le Chatelier's principle predicts that the equilibrium will shift to increase the
concentration of products.
*Increasing the rate of the forward reaction will mean an increase in products.
*So some sulfur dioxide or oxygen is used to produce sulfur trioxide.Equilibrium shifts to the right. That is, when a new equilibrium is reached there will
be more product than before.
25.F2----->2F
Initial pressure of F2is 0.5atm.
Final pressure of F2=(0.5-x) and pressure of F is 2x.
For F2 final pressure is taken 0.5 because the value of Kp is much less.
Kp=2.4×10^-8
Kp=[F]^2/[F2]
2.40×10^-8=4x^2/0.5
x =5.46×10^-4
2X =0.546×10^-4×2=1.09×10^-4atm
26.2NOCl---------->2NO+Cl2
K1=[NO]^2[Cl2]/[NOCl]^2
NO +1/2Cl2--------->NOCl
K2=[NOCl]/[NO][Cl2]^1/2
To compare K1,K2
K2=K1
K2=0.51
=1.4
27.(a)If a catalyst is added to a reaction, both the forward and reverse reaction rates will be
increased. If both rates are increased then the concentrations of the reactants and products
will remain the same. This means that a catalyst has no effect on the equilibrium position.
However, a catalyst will affect how quickly equilibrium is reached. This is very important in
industry where the longer a process takes, the more money it costs. So if a catalyst reduces
the amount of time it takes to form specific products, it also reduces the cost of production.
(b)
*Le Chatelier's principle predicts that the equilibrium will shift to increase the
concentration of products.
* Increasing the rate of the forward reaction will mean an increase in products.
* So some PCl5 is used to produce PCl3 and Cl2.
(c)
Adding an inert gas reduces the mole fraction of PCl5, PCl3 and Cl2. However, the effect of reducing the mole fraction of the products is greater than that of of the reactants. Therefore, the favoured reaction is one that increases the molar fraction of the products, ie. the forward the reaction.
(d)Le Chatelier's principle predicts that the equilibrium will shift to increase the
concentration of reactants.
* Increasing the rate of the reverse reaction will mean an increase in reactants.
* So some PCl3 and Cl2 would change back to PCl5 to restore
equilibrium.
* Equilibrium shifts to the left. That is, when a new equilibrium is reached there will
be less product than before.
This means there will be more PCl3 than the first equilibrium.