[ArtVandelay_45]

I usually don't do this, but I'm having unusual trouble on some of these questions.

1. At 425 oC, Kp = 4.18 x 10-9 for the reaction 2HBr (g) ><H2(g) + Br2(g)


In one experiment, 0.20 atom of HBr(g), 0.010 atm of H2(g), and 0.010 atm of Br2(g) are introduced into a container. Is the reaction at equilibrium? If not, in which direction will it proceed? Explain your reasoning.


My answer:

Keq= [C]^c [D]^d / [A]^A

= (.10) (.10) / (.20)2

=1/4

Keq is less than one and therefore the EqM lies to the left.

 

2. In an analysis of interhalogen reactivity, 0.500 mol of ICl was placed in a 5.00 L flask

and allowed to decomposed at a high temperature. 2 ICl(g)>< I2(g) + Cl2(g). Calculate the equilibrium concentrations of I2, Cl2 and ICl. (Kc = 0.110 at this temperature)


My answer (so far, not done):

             I           C           E

[ICl]   .10        -2x         .1-2x

[I2]      0           +x           x

[Cl2]   0            +x           x

Kc= [I2][Cl2] / [ICl]^2

0.110= x2 / (.1-2x)2

(0.1 - 2x) (square root of .110) =X

.03317 -.663X=X

1.663 X=.3317

5=X

Don't know where to go from here...

 


3. When 0.0150 mol of NH3(g) and 0.015 mol of O2(g) are placed in a 1.00 L container the N2 concentration at equilibrium is 1.96 x 10-3 mol/L. Calculate Kc for the reaction at this


temperature. 4NH3(g) + 2 O2(g) >< 2 N2(g) + 6 H2O(g)

My answer (so far, not done):

Kc= [N2]2 [H2O]6  /  [ NH3]6 [O2]2

               I                 C                       E

[NH3]      0.0015       -1.96 x 10-3        -4.6 x 10-4

[O2]        0.0015       -1.96 x 10-3         -4.6 x 10-4

[N2]        0                + 1.96 x 10-3     1.96 x 10-3

[H2O]     0                 + 1.96 x 10-3     1.96 x 10-3

Kc= (1.96 x 10-3) 2 (1.96 x 10-3) / (-4.6 x 10-4)4 (4.6 x 10-4)2 = 3.65 x 1015 mol/L

 


4. At elevated temperatures, SbCl5 gas decomposes into SbCl3 gas and Cl2 gas as shown by


the following equation: SbCl5(g) « SbCl3(g) + Cl2(g)


(a) An 89.7 gram sample of SbCl5 (molecular weight 299.0) is placed in an evacuated 15.0


litre container at 182ºC.

1. What is the concentration in moles per litre of SbCl5 in the container before any decomposition occurs?


2. What is the pressure in atmospheres of SbCl5 in the container before any decomposition occurs?


(b) If the SbCl5 is 29.2 percent decomposed when equilibrium is established at 182ºC,


calculate the value for either equilibrium constant Kp or Kc, for this decomposition reaction. Indicated whether you are calculating Kp or Kc.


(c) In order to produce some SbCl5, a 1.00 mole sample of SbCl3 is first placed in an empty 2.00 litre container maintained at a temperature different from 182ºC. At this temperature, Kc, equals 0.117. How many moles of Cl2 must be added to this container to


reduce the number of moles of SbCl3 to 0.700 mole at equilibrium?


My answer:

1)

(89.7 g SbCl5)(1 mol/299.0 g)  =  0.300 mol SbCl5

[SbCl5] i  =  0.300 mol/15.0 L  =  0.200 M

(2)

T = 182° C + 273 = 455K

P  =  nRT/V  =  (0.300 mol)(0.0821 L-atm/K-mol)(455 K)/15.00 L  =  0.747 atm




--------------------------------------------------------------------------------



b)

[SbCl3]  =  [Cl2]  =  (0.0200 mol/L)(0.292)  =  5.84 x10-3 M

[SbCl5]  =  (0.0200 mol/L)(0.708)  =  1.42 x10-2 M


Kc  =  [SbCl3][Cl2]/[SbCl5]  =  [(5.84 x10-3)2]/[1.42 x10-2]  =  2.41 x10-3

 


--------------------------------------------------------------------------------

c)

K = ([SbCl3][Cl2]) ¯ [SbCl5] = 0.117

Equilibrium concentrations:

[SbCl5] = (1.00 - 0.70) mol / 2.00 L = 0.15 M

[SbCl3] = 0.700 mole / 2.00L = 0.350 M

[Cl2] = x

Kc = [(0.350) (x)] ÷ (0.15) = 0.117


x = [Cl2] = 0.50 M

Moles Cl2 at equilibrium = 0.050 mol L x 2.00 L = 0.10 mol


Moles Cl2 needed to make 0.300 mol SbCl3 into SbCl5 = 0.30 mol


Moles Cl2 that must be added = 0.40 mol

5. 2 H2S(g) ® 2 H2(g) + S2(g)


When heated, hydrogen sulfide gas decomposes according to the equation above. A 3.40 g sample of H2S(g) is introduced into an evacuated rigid 1.25 L container. The sealed container is heated to 483 K, and 3.72´10–2 mol of S2(g) is present at equilibrium.


(a) Write the expression for the equilibrium constant, Kc, for the decomposition reaction represented above.


(b) Calculate the equilibrium concentration, in mol·L-1, of the following gases in the container at 483 K.


(i) H2(g)


(ii) H2S(g)


(c) Calculate the value of the equilibrium constant, Kc, for the decomposition reaction at 483 K.


(d) Calculate the partial pressure of S2(g) in the container at equilibrium at 483 K.


(e) For the reaction H2(g) + 1/2 S2(g) ® H2S(g) at 483 K, calculate the value of the equilibrium constant, Kc.

My answer:


(a) Kc =[H2]2[S2] / [H2S]2


(b) (i) (3.72810–2 mol S2 / 1.25 L) x (2 mol H2 / 1 mol S2) =  5.95×10-2 M H2


(ii) (3.40g H2 S X 1 mol/34g) -(3.72 x 10-2 mol S2 x 2mol H2S / 1 mol S2) / 1.25 L = 2.05 x 10-2 M H2S

(c) Kc= ((5.952 x 10-2)2 x (3.72 X 10-2/1.25)) / (0.02048)2=.251




(d) PV = nRT = 1.18

(e) K’c = square root of 1/Kc=2




6. Le Chatelier's Principle is ultimately related to the rates of the forward and reverse steps in a reaction. Using this idea explain why;

a) increasing reactant concentration shifts the equilibrium position to the right but does not change the equilibrium constant.

b) decreasing volume shifts the equilibrium toward fewer moles of gas but does not change the equilibrium constant.

c) increase in temperature shifts the equilibrium position of an exothermic reaction towards reactants and also changes the equilibrium constant.

My answer:

Don't know where to start

7. 2 CH4(g) C2H2(g) + 3H2(g) (exothermic)


Predict the effect of the following changes on the moles of CH4 present at equilibrium:


a) increasing the volume of the container

b) increasing pressure in the container by adding an inert gas

c) adding some extra H2

MY answer:

Don't know where to start

[ArtVandelay_45]

17 views and no replies? Are these questions that hard?