chap 14

Question 1

what is chemical equilibrium

Answer 1

concentrations dont change BUT are technically going back and forth at same rate

Question 2

what is Q?

Answer 2

numerical value of the mass action expression for any values of concentrations or pressures of the reactants and the products. its equal to Keq formula but but Keq is at equilibrium and Q is not at equilibrium.

Question 3

what does RICE stand for?

Answer 3

reaction, initial, change, equilibrium

Question 4

what does an intermediate value of K mean? Kc= 24

Answer 4

comparable amounts of products and reactants at equilibrium

Question 5

what does a very large K mean? Kc= 3 x 10⁸¹

Answer 5

favors formation of products

Question 6

what does a very small K mean? Kc= 3 x 10 -⁸¹

Answer 6

favors reactants. not much product formed at equilibrium

Question 7

when Q<K the reaction will proceed in the _____ direction 🔺️G

Answer 7

forward

Question 8

when Q>K the reaction will proceed in the _____ direction. 🔺️G

Answer 8

reverse

Question 9

when Q=K the reaction is at _______. 🔺️G = 0

Answer 9

equilibrium

Question 10

🔺️Grxn<0 is _______ when Q<Keq. rxn proceeds(products favored)

Answer 10

spontaneous

Question 11

🔺️Grxn>0 is ________ when Q>Keq. reverse rxn proceeds(reactants favored)

Answer 11

non-spontaneous

Question 12

what does 🔺️G°rxn measure

Answer 12

hpw far the rxn is from equilibrium under standard conditions

Question 13

what is Homogeneous equilibria

Answer 13

reactants and products in the same phase

Question 14

what is Heterogeneous equilibria

Answer 14

reactants and products in different phases. solid and liquid and aqeuous do not get used to find Keq

Question 15

what is Le Châtelier’s Principle

Answer 15

Factors that will change the relative rates of forward/reverse reactions or change the
value of Q compared to Keq will cause a shift in the position of equilibrium.

Question 16

If total moles of gas does not change, then equilibrium ______ shift

Answer 16

will not

Question 17

decreasing the volume increases the pressure, which also increases M, causing reaction to shift to the _____(think syringe)

Answer 17

right

Question 18

according to Le Châtelier’s Principle if you Compress the reaction mixture
(decrease V or increase P) the rxn will shift ______

Answer 18

right

Question 19

according to Le Châtelier’s Principle if you Expand the reaction mixture
(increase V or decrease P) the rxn will shift ______

Answer 19

left

Question 20

Increasing temperature adds heat to the right side of equation(exothermic); reaction shifts ____

Answer 20

left

Question 21

endothermic: increase temp-shift ____
decrease temp-shift ____
exothermic: increase temp-shift _____
decreased temp-shift ____

Answer 21

right, left, left, right

Question 22

what does adding a catalyst do to rxn?

Answer 22

Systems reach equilibrium faster because activation energy (Ea) lowered, but NO CHANGE in K or G or position of equilibrium.

Question 23

CO(g)+H2O(g)=CO2(g)+H2(g)
The water-gas shift reaction is an important source of hydrogen. The value of Kc for the reaction below is is 5.1 at 700 K. Calculate the equilibrium concentrations of the four gases if the initial concentration of each of them is 0.0954 M.

Answer 23

Kc=([CO2][H2])/([CO][H2O])
CO(g) + H2O(g) = CO2(g) + H2(g)
0.0954 0.0954 0.0954 0.0954
-x -x +x +x
0.0954-x 0.0954-x 0.0954+x 0.0954+x
5.1=(0.0954+x)(0.0954+x)
(0.0954-x)(0.0954-x)
5.1=(0.0954+x)²
(0.0954-x)
square root both and solve for x
x= 0.0368 PLUG IN X
0.0954-0.0368= 0.0586 M CO and H2O
0.0954+0.0368= 0.1322 M CO2 and H2

Question 24

1/4A(s) + 3B(g) = 2C(g)
For a reversible reaction the concentration of [C] at equilibrium is 0.0910 M, and K = 3.5 × 10^–3. Find [B] at equilibrium.

Answer 24

K= ([C]²)/([B]³) A doesnt count cause (s)
3.5 × 10^–3= (0.0910)²/[B]³
B³= (0.0910)²/3.5 × 10^–3
B=1.33 M

Question 25

Calculate the value of the equilibrium constant K at a temperature of 298 K for each value of ΔGº. ΔGº = 7.50 kJ/mol and ΔG° = -9.00 kJ/mol

Answer 25

G° = –RT lnKeq 7.50= -(0.008314)(298) lnKeq Keq=7.50/-2.478 K= e-³.⁰²⁷= 0.0485 same operation for other one= 37.8

Question 26

A sealed chamber contains an equilibrium mixture of NO2(g) and N2O4(g) at 15°C. Their partial pressures are 𝑃NO2 = 0.34 atm and 𝑃N2O4 = 0.80 atm. What is the value of Kp for the following reaction under these conditions? 2NO2(g) ⇌ N2O4(g)

Answer 26

Kp=Keq Kp= (0.80)/(0.34)² = 6.92 unitless

Question 27

Calculate the equilibrium partial pressures for each reactant and product: CO(g) + H2O(g) ⇌ CO2(g) +H2(g) 0.50 atm 0.25 atm 1.50 atm 2.00 atm Kp = 10.0 at 400 °C

Answer 27

Q= ((1.5)(2))/((0.5)(0.25))=24 Keq is equal to 10 so all values go down to ten +x +x -x -x 0.5+x 0.25+x 1.5-x 2.0-x 10=((1.5-x)(2-x))/((0.5+x)(0.25+x))

Question 28

when doing: 10= (2x²)/((0.5+X)(0.25+X)) and our Keq is 1.0 x 10^(-5) x is very small (Keq= 10^(-5)) so we can ignore these x's(X) cannot do top one bc we arent subtracting or adding to it. when do we do?

Answer 28

use this when Keq is < 10^(-3)

Question 29

Use the G°f values listed in Appendix 4 to calculate G°rxn and the value of Keq for the formation of 1 mole of NO2 at 298 K: NO(g) + 1/2 O2(g) ⇌ NO2(g) ∆𝐺(kJ/mol) = 86.6 0.0 51.3

Answer 29

G°rxn = E[(n)(G°fno2)] - [(n)(NOf) + (n)(O2f)] (1)(51.3) - (1)(86.6)+ 1/2(0.0)= -35.3 kJ/mol at eq: G°rxn = -RTln(Keq) -35.3 = -(.008314)(298)ln(Keq) 14.25 = ln(Keq) ---> Keq= e¹⁴.²⁵ ----> Keq= 1.5x 10⁶ products favored

Question 30

For the chemical equilibrium A + B = C, the value of the equilibrium constant K is 10. What is the value of the equilibrium constant for the reaction 2C = 2A + 2B?

Answer 30

doing reverse rxn reverses the K. 1/10 but then since its double (1/10)² so 1/100= 0.01

Question 31

The forward rate constant, kf, and reverse rate constant, kr, for a chemical reaction are not equal. Which statement below must be true? a) kf and kr will remain unequal, but the rates will become equal owing to concentration changes. b) kf and kr will remain unequal, but the rates will become equal owing to temperature changes. c) kf and kr will become equal as equilibrium is approached owing to temperature changes.

Answer 31

a.) kf and kr will remain unequal, but the rates will become equal owing to concentration changes.