Chem 1A – CLAS – Ch 6 – Key

1. Consider the following reaction: N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g)

a. Write the equilibrium expression (K and Kp)

K  = [ NCl3 ]2

[ N2 ] [Cl2 ]3 and Kp =

( PNCl 3 ) 2( PN 2) ( PCl 2 )3

b. What is the value of Kp at 25°C if the equilibrium pressures for N2, H2 and NH3 were 1.8 atm, 0.5 atm and 0.017 atm respectively.

Kp = (0.017 ) 2

(1.8 ) (0.50 )3 = 0.0013

c. What is the value of K at 25°C? Kp = K(RT)Δn

K = (0.0013)(0.08206 x 298)(2−4)

= 0.77

d. Does the reaction favor reactants or products? When K < 1 the reactants are favored

2. Which reaction does the K = Kp? a. P4 (s) + 6 Cl2 (g) ⇌ 4 PCl3 (g)b. H2O (l) ⇌ H2O (g)c. H2 (g) + Cl2 (g) ⇌ 2 HCl (g) ⇒ moles of gas reactant = moles of gas productd. C3H8 (g) + 5 O2 (g) ⇌ 3 CO2 (g) + 4 H2O (g)

3. Given the following information:H2(g) + I2(g) ⇌ 2 HI(g) K = 54.0N2(g) + 3 H2(g) ⇌ 2 NH3(g) K = 96.0

Determine the value of the equilibrium constant for the following reaction:2 NH3(g) + 3 I2(g) ⇌ 6 HI(g) + N2(g)

Reaction 1 needs to be multiplied by 3 ⇒ 3 H2(g) + 3 I2(g) ⇌ 6 HI(g) K = 54.03

Reaction 2 needs to be flipped ⇒ 2 NH3(g) ⇌ N2(g) + 3 H2(g) K = 96.0-1

Add reactions together ⇒ 2 NH3(g) + 3 I2(g) ⇌ 6 HI(g) + N2(g) K = (54.03)(96.0-1) = 16404. Consider the following reaction:

UO2 (s) + 4HF (g) ⇌ UF4 (g) + 2H2O (g) Kp = 11If you fill a 5 L container with 15 g of UO2, 1.1 atm of HF, 0.75 atm of UF4 and 5.9 atm of H2O, what will be the mass of the solid UO2 at equilibrium?a. 15 gb. > 15 gc. < 15 gd. not enough information

5. Consider the following reaction: N2 (g)  +  O2 (g) ⇌ 2  NO (g)    Kp = ? Initially the partial pressures of N2 and O2 are 1 atm and 3 atm respectively. At equilibrium the pressure of NO is 1.5 atm. What is the Kp for this reaction?

N2 (g) O2 (g) ⇌ 2  NO (g)

I 1 atm 3 atm 0C - x - x +2xE 1 atm – 3 atm – 2x

Qp = (P UF4 ) (P H2O ) 2

(P HF ) 4 =

(0.75 ) (5.9 )2(1.1 ) 4

= 17.8

Since K < Q the reaction will go backwards thereby increasing the mass of the UO2

Since the P of NO at equilibrium is 1.52x = 1.5 => x = 0.75So at equilibrium the

PN2 and PO2 are 0.25 atm and 2.25 atm respectively

Kp= (P NO )2

(P N2 ) (P O2 )=(1.5)2

(0.25)(2.25) = 4

x x

6. Pure PCl5 is introduced into an empty rigid 5-L flask at 24 °C with a pressure of 0.45 atm.  The PCl5 is heated to 90 °C upon which it decomposes into solid P and gaseous Cl2. 

2 PCl5 (g) ⇌ 2 P (s) + 5 Cl2 (g)At equilibrium the total pressure is measured to be 0.93 atm.  What is the equilibrium constant (Kp) at 90 °C? What percentage of the PCl5 decomposed? How many grams of P are in the flask at equilibrium?

2 PCl5 (g) ⇌ 2 P (s)

5 Cl2

(g)I 0.55 atm 0 0C - 2x + 2x +5xE 0.55 atm –

2x 2x 5x

7. At a certain temperature the partial pressures of an equilibrium mixture of N2O4 (g) and NO2 (g) are 0.34 atm and 1.2 atm respectively.

N2O4 (g) ⇌ 2 NO2 (g) The volume of the container is doubled. Find the new partial pressures when the equilibrium is re-established.

Skip for Kunkel’s class8. Consider the following reaction: 

2 NOBr (g) ⇌ 2 NO (g) + Br2 (g)     Kp = 34 An unknown pressure of NOBr is put into a rigid container at 40 °C, equilibrium is reached when 86 % of the original partial pressure of NOBr has decomposed.  What was the initial pressure of the NOBr?  What is the total pressure in the flask at equilibrium?

2 NOBr (g)

⇌ 2 NO (g)

Br2 (g)

I x 0 0C - 0.86x + 0.86x +0.43

xE 0.14x 0.86x 0.43x

9. At a particular temperature Kp = 5.7x10-8 for the following reaction:  2 C (s) + 3 H2 (g) ⇌ C2H6 (g)

What is the pressure of C2H6 at equilibrium if initially there's 5 g of C and 3 atm of H2?

2 C 3 H2 ⇌ C2H6

Using the combined gas law⇒ P 1V 1n 2T 2 =

P 2 V 2n2T 2

Where n and V are constant ⇒ (0.45 atm )(24+273 K )

=(P 2 )

(90+273 K )P2 = 0.55 atm

Ptotal = PPCl5 + PCl2

0.93 = (0.55 – 2x) + (5x)x = 0.13

PPCl5 = 0.55 – 2(0.13) = 0.29PCl2 = 5(0.13) = 0.65

Kp = (0.65)5(0.29)2

Kp = 1.38

% decomposed = amount decomposed

initial amount x100

Use K to solve for x ⇒ 34 = (0.86x)2(0.43x)(0.14x)2 ⇒ x = 2.2 atm ⇒

the initial pressure of NOBr is 2.2 atmThe total pressure in the flask at equilibrium

Ptotal = 0.14x + 0.86x + 0.43x = 3.1 atm

Since Kp is so small we can make the assumption that x is negligible with respect to a non-zero value

Kp = (P C2H6  )(P H2  )3

5.7x10-8 = x(3)3 

(g) (g) I 5g 3 atm 0C - 2x - 3x +xE 5g –

x3 – 3x x

10. Consider the following endothermic reaction at equilibrium:  CCl4 (g) ⇌ C (s) + 2 Cl2 (g)

a. Is heat absorbed or released as the reaction goes forward? absorbedb. Which way will the reaction shift if the temperature is increased? rightc What happens to K as the temperature is increased? increasesd. Which way will the reaction shift if the partial pressure of CCl4 is increased? righte. Which way will the reaction shift if you add more C? no shiftf. Which way will the reaction shift if you remove Cl2? rightg. Which way will the reaction shift if you compress the system? lefth. Which way will the reaction shift if you add a catalyst? no shift

Equilibrium Expression jA + kB ⇌ lC + mD

K = [C ] l [ D ]m

[ A ] j [ B ]kKp =

PCl PD

m

PAj PB

k

Pure solids and liquids are given the value of 1 in the expression

Where does the equilibrium lie?K > 1 ⇒ Products are favored K < 1 ⇒ Reactants are favored

Since reversible reactions can be written in either direction or balanced with any multiple of coefficients you must consider how this affects the equilibrium constant (K or Kp) 1. If you flip a reaction take the reciprocal of the constant

ex: if K=5 for 2A ⇌ B then K= 15 for B ⇌ 2A

2. If you multiply the reaction coefficients raise the constant to the same multipleex: if K=5 for 2A ⇌ B then K=5½ for A ⇌ ½B

3. If you add two or more reactions take the product of the constantsex: if K=5 for 2A ⇌ B and if K= 8 for B ⇌ 3C thenK = (5)(8) = 40 for 2A ⇌ 3C

To determine the direction a reaction must proceed to get to equilibriumK > Q shifts to the right (causes Q to increase until K=Q)K < Q shifts to the left (causes Q to decrease until K=Q)K = Q reaction is at equilibrium so no shift

Since Kp is so small we can make the assumption that x is negligible with respect to a non-zero value

Kp = (P C2H6  )(P H2  )3

5.7x10-8 = x(3)3