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Edit View History Bookmarks People Window Help Watch Free Nin/Tuck (20031-x EvaluationKIT . student feedb_x·A Hamenace . CHE 101C07 intr-x-m MasteringChemistry: Chap 10 exam part 2 Secure https://session.masteringchemistry.com/myct/itemView?assignmentProblemiD-89347948 ap 10 exam part 2Item 1 Time Remaining: 00:39:53 Hide Item 1 Part A -2 NH3 (g) 46 kJN2 (g)3 H2(g) For the reaction at equilibrium given above, give 2 ways that you can shift the equibrium to the left Essay answers are limited to about 500 words (3800 characters maximum, including spaces). 3800 Character(s) remaining Submit My Answers Give Up Part B-2 NH3 (g)+46 kJN2 (g)+3 H2(g) For the reaction at equilbrium given above, calculate the numeric value of Kc if the concentrations at equilibrium areo Essay answers are limited to about 500 words (3800 characters maximum, Including spaces). 3800 Character(s) remaining Submit My Answers Give Up Part C-Are all reversible reactions at equilibrium? Explain why or why not Limit your answer to 3-5 sentences. In order to receive credit, please answer in complete sentences

Explanation / Answer

a.

shift to the left implies:

increase of NH3 ...

a) addition of either N2 and H2, will favour plenty fo products, so reverse reactino will occur

b) removal of heat, since this is exothermic, requires heat, if we refrigerate the system and decrease T, this will favour more NH3 formation

B)

numeric value of Kc:

dG = exp(-DG/(RT))

K= exp(-91800/(8.314*298))

K = 8.097*10^-17

C)

Technically speaking, all reactions are in equilbirium, since their rate of reaction are the same, i.e.

rate of products = rate of reactants

D)

if K = 10^-12

this implies

K = products / reactants

reactants >> products

thereofre, this favour reactants mostly

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