15. The reaction rates of many spontaneous reactions are actually very slow. Whi

Question

15. The reaction rates of many spontaneous reactions are actually very slow. Which of these statements is the best explanation for this observation? A) Kp for the reaction is less than one. B) The activation energy of the reaction is large. C) AGo for the reaction is positive. D) Such reactions are endothermic. E) The entropy change is negative. 16. Ozone (O,) in the atmosphere can reaction with nitric oxide (NO): 0)(g) + NO(g) NO2(g) + O2(g). Calculate the G) for thi A) 1020 kJ/mol s reaction at 25°C. (AH=-199 kJ/mol. AS-4. I J/K-mol) CF. 2.00 × 101 kJ/mol D) E) -1.42 x 103 kJ/mol -198 kJ/mol 17. Hydrogen peroxide (H:O2) decomposes according to the equation H:O:(I) H20(I) + 1402(g). Calculate K, for this reaction at 25°C. (AH98.2A/mol,AS-70.1 J/K mol) A) 1.3×10-21 B) 20.9 C) 3.46 x 10 D) 7.5x 1020 E) 8.6 x 10 K. for the auto-ionization of water, HO(1)HTaq) +0H (aq), is! 0x1 the signs (+/-) of ASo and AH for the reaction at 25 C? 18. -What are 19. Which of the following is consistent with a reaction that proceeds spontaneously in the forward direction? C) @c0. Q > K

Explanation / Answer

15)

it is because , the activation energy is large

so

the answer is option B

16)

we know that

dGo = dHo - TdSo

given

temperature = 25 + 273 K = 298 kelvin

so using given values

we get

dGo = ( -199 x 1000 ) - ( 298 x -4.1)

dGo = -199000 + 1221.8

dGo = -197778

dGo = -197.778 x 1000 J/mol

dGo = -197.778 kJ/mol

so

the answer is E) -198 kJ/mol

17)

now

dGo = dHo - TdSo

dGo = (-98.2 x 1000) - ( 298 x 70.1)

dGo = -119089.8 J/mol

now

we know that

dGo = -RT lnKp

so

-119089.8 = -8.314 x 298 x lnKp

Kp = 7.5 x 10^20

so

the answer is D) 7.5 x 10^20

18)

we know that

the entropy order is

gas > ions > liquids > solids

so

individual ions have more entropy then a liquid solution

now

H20 (l) ---> H+ + OH-

we can see that

there is an increase in entropy

so dSo > 0

also auto ionization of water is endothermic , dH > 0

so

the answer is A) dSo= (+) , dHo = (+)

19)

we know that

for a reaction to be spontaneous , dG < 0

and

we know that

if Q < K , the reaction moves in the forward direction

if Q = K , no change , reaction at equilibrium

if Q > K , the reaction moves in the backward direction

so

Q < K for the reaction to proceed spontaneously in the forward direction

the answer is E) dG < 0 and Q &lt%3C/strong>

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