6. Consider the following reaction at constant P. Use the information here to de

Question

6. Consider the following reaction at constant P. Use the information here to determine the value of Ssurr at 355 K. Predict the spontaneity of the reaction at this temperature. 2 NO(g) + O2(g) 2 NO2(g) H = -114 kJ

A) Ssurr = +114 kJ/K, reaction is spontaneous

B) Ssurr = +114 kJ/K, reaction is not spontaneous

C) Ssurr = +321 J/K, reaction is spontaneous

D) Ssurr = -321 J/K, reaction is not spontaneous E) Ssurr = +321 J/K, it is not possible to predict the spontaneity of this reaction without more information.

7. Consider a reaction that has a positive H and a positive S. Which of the following statements is TRUE?

A) This reaction will be spontaneous only at high temperatures.

B) This reaction will be spontaneous at all temperatures.

C) This reaction will be nonspontaneous at all temperatures.

D) This reaction will be nonspontaneous only at high temperatures.

E) It is not possible to determine without more information.

8. Predict the sign of S in the system for each of the following processes:

a. CO2(s) CO2(g) (sublimation of dry ice)

b. CaSO4(s) CaO(s) + SO3(g)

c. N2(g) + 3 H2(g) 2 NH3(g)

d. I2(s) I2(aq)(dissolution of iodine in water)

Explanation / Answer

6. For the given reaction, the reaction entropy (surr) would be,

D) dSsurr = -321 J/K, the reaction is non-spontaneous

dSsurr = dH/T = -114 x 1000/355 = -321 J/K

7. For a reaction with +ve dH and +ve dS, the reaction will be spontaneus at,

A) this reaction will be spontaneous only at hight temperatures.

8. Sign of dSsys,

a. Sublimation increases entropy (solid-to-gas), so dSsys would be +ve.

b. We see an increase in entropy, two molecules (one solid and other gas) is formed from 1 molecule of solid CaSO4, so increase in entropy, sign would be +ve for dSsys.

c. 4 molecules going to 2 molecles, decrease in entropy, dSsys would be -ve

d. I2 solid to liquid would be with an increase in entropy. So dSsys woud be +ve.

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