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Jet001 [13]
3 years ago
7

(c) Assume you have an equilibrium mixture of [A], [B], and [C] at 298K and that the

Chemistry
1 answer:
djyliett [7]3 years ago
3 0

Answer:

Explanation:

1. The amount of CaCO3 must be so small that  

P

CO

2

 is less than KP when the CaCO3 has completely decomposed. In other words, the starting amount of CaCO3 cannot completely generate the full  

P

CO

2

 required for equilibrium.

3. The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants’ side.

5. No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.

7. Add N2; add H2; decrease the container volume; heat the mixture.

9. (a) ΔT increase = shift right, ΔP increase = shift left; (b) ΔT increase = shift right, ΔP increase = no effect; (c) ΔT increase = shift left, ΔP increase = shift left; (d) ΔT increase = shift left, ΔP increase = shift right.

11. (a)  

K

c

=

[

CH

3

OH

]

[

H

2

]

2

[

CO

]

; (b) [H2] increases, [CO] decreases, [CH3OH] increases; (c), [H2] increases, [CO] decreases, [CH3OH] decreases; (d), [H2] increases, [CO] increases, [CH3OH] increases; (e), [H2] increases, [CO] increases, [CH3OH] decreases; (f), no changes.

13. (a)  

K

c

=

[

CO

]

[

H

2

]

[

H

2

O

]

; (b) [H2O] no change, [CO] no change, [H2] no change; (c) [H2O] decreases, [CO] decreases, [H2] decreases; (d) [H2O] increases, [CO] increases, [H2] decreases; (f) [H2O] decreases, [CO] increases, [H2] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.

15. Only (b)

17. Add NaCl or some other salt that produces Cl− to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s).

19. (a)

Hope this helps :)

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