Question

16. The only factor that changes the equilibrium constant of a chemical reaction is: A) pressure B) concentration C) volume D) temperature 17. Determine the effect of increased pressure on the system in the following equilibrium: C(graphite)+S2(g) CS2 (g) A) The equilibrium position SHIFT to the right B) The equilibrium position SHIFT to the left C) does not change the position of equilibrium D) The Le Chatelet principle does not apply to the system. 18. If the acetic acid dissociation constant is Ka 1.8 X 10-5 ,and its concentration before dissolution is( 0.34M), the concentration of [H +] in an acid solution is equal to the following equation CH COOH CHCOO + H C) 3.52X10 B) 2.47X103 D) 2.47X10 A) 3.52X103 19. If K 0.45 at a (150°C) for the equilibrium reaction: N2 (g)O2 (g) 2NO The Ke of the reaction is: B) 0.54 C) 0.69 D) 0.96 A) 0.45

Answer 1

16)

Value of Kc depends only on temperature

Answer: D

17)

Increasing pressure will shift the reaction in a direction which have lesser gaseous molecules as per Le chatelier Principle

Here, number of gaseous molecules is same on both side

So equilibrium will not be effected

So, No effect on equilibrium

Answer: C

18)

CH3CO2H dissociates as:

CH3CO2H -----> H+ + CH3CO2-

0.34 0 0

0.34-x x x

Ka = [H+][CH3CO2-]/[CH3CO2H]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.8*10^-5)*0.34) = 2.474*10^-3

since c is much greater than x, our assumption is correct

so, x = 2.474*10^-3 M

So, [H+] = x = 2.474*10^-3 M

Answer: B

19)

T= 150.0 oC

= (150.0+273) K

= 423 K

Δ n = number of gaseous molecule in product - number of gaseous molecule in reactant

Δ n = 0

use:

Kp= Kc (RT)^Δn

0.45 = Kc *(0.08206*423.0)^(0)

Kc = 0.45

Answer: A