Calculate the concentration of NH_3 gas at equilibrium if a 1.50-L reaction vess

Question

Calculate the concentration of NH_3 gas at equilibrium if a 1.50-L reaction vessel initially contains 6.00 mol of NH_3, 9.00 mol of H_2 S, and 31.0 mol of NH_4 HS. At the temperature of the reaction, K = 5.5. NH_3(g) + H_2 S(g) NH_4 HS (s) a) 4.04 M b) 3.88 M c) 1.25 M d) 0.087 M e) 0 M Use the following to answer questions 12-14: Given the equation 2A(g) 2B (g) + C(g). At a particular temperature, K = 1.6 times 10^4 If you mixed 5.0 mol B, 0.10 mol C, and 0.0010 mol A in a one-liter container, which direction would the reaction initially proceed? a) To the left. b) The above mixture is the equilibrium mixture. c) To the right. d) Cannot tell from the information given. e) None of these (A-D). Addition of chemical B to the reaction mixture above, at equilibrium, will a) cannot be determined b) have no effect c) cause [A] to increase d) cause [C] to increase e) none of the above At a higher temperature, K = 1.8 times 10^-5. Placing the equilibrium mixture in an ice bath (thus lowering the temperature) will a) have no effect b) cause [A] to increase c) cause [B] to increase d) cannot be determined e) none of the above

Explanation / Answer

Concentration= moles/unit volume (L)

Concentrations of gases (NH3)= 6/1.5= 4M and H2S= 9/1.5= 6M

The reaction is NH3(g)+ H2S(g)ß-> NH4HS(s)

There are mole mole of

At equilibrium [NH3]= 4-x, [H2S]= 6-x

KC= [NH4HS]/[NH3] [H2S] , since the activity of [NH4HS] is unity.

KC= 1/[NH3] [H2S]= 1/(4-x)*(6-x)= 5.5

When solved using excel, x= 3.913, hence [NH4] at equilibrium= 4-3.913=0.087M ( d is correct)

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