The following equilibrium represents reversible reaction of obtaining the ammoni
ID: 516015 • Letter: T
Question
The following equilibrium represents reversible reaction of obtaining the ammonium N_2(g) + 3 H_2(g) 2 NH_3(g) + heat a) What change in equilibrium do you expect when you add N_2 to reaction mixture? Explain your answer in terms of LeChatelier's principle. b) What change in equilibrium do you expert when you decrease the temperature of the reaction mixture? Explain your answer in terms of LeChatelier's principle. Evaluating the Equilibrium Constant for the Reaction of Iron(III) Ion with Thiocyanate Ion a) Determine an equilibrium constant of the following reaction FeNCS^2+ (aq blood red) doubleheadarrow Fe^3+ (aq. pale yellow) + SCN (aq. colorless) b) The initial concentrations of Fe^3+ and SCN are 0.040M and 0.800M. respectively It was determined that the equilibrium concentration of FeNCS^2+ is equal to 0.040M Calculate equilibrium constant of reaction.Explanation / Answer
7 a) There will not be any change in equilibrium because 1 mol of Nitrogen is enough to react with 3 moles of Hydrogen to produce 2 moles of ammonia. So even if you add excess of nitrogen it will not affect the equilibrium. The excess nitrogen will remain in the reaction mixture as such.
b) When we increase the temperature the reaction mixture gets more heat and there by it leads to the formation of nitrogen and hydrogen (because it is true that NH3 and heat would produce N2 and H2) that means the equilibrium shifts left. In the same way when we reduce temperature the formation of Ammonia will be more, that means the equilibrium constant increases for that forward reaction.
8 a) The equilibrium constant (K) for the given reaction is {[Fe3+].[SCN-]}/[FeNCS2+], where [Fe3+], [SCN-], [FeNCS2+] are the concentrations of Fe3+, SCN-, FeNCS2+ respectively.
b) From the previous formula to calculate equilibrium constant we can substitute the initial concentrations of reactants and equilibrium concentration of product. Finally we get (0.040M)(0.800M)/(0.040M)= 0.8M as equilibrium constant.
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