A mixture of 0.10 mol{ m mol} of NO{ m NO}, 0.050 mol{ m mol} of H2{ m{H}}_2, an

Question

A mixture of 0.10 mol{ m mol} of NO{ m NO}, 0.050 mol{ m mol} of H2{ m{H}}_2, and 0.10 mol{ m mol} of H2O{ m{H}}_2 { m{O}} is placed in a 1.0-L{ m L} vessel at 300 K{ m K}. The following equilibrium is established:

Part A

Express your answer using two significant figures.

Part B

Express your answer using two significant figures.

Part C

Part D

Express your answer using two significant figures.

A mixture of 0.10 mol{rm mol} of NO{rm NO}, 0.050 mol{rm mol} of H2{rm{H}}_2, and 0.10 mol{rm mol} of H2O{rm{H}}_2 {rm{O}} is placed in a 1.0-L{rm L} vessel at 300 K{rm K}. The following equilibrium is established: Part A Calculate the equilibrium concentration of H2 K_c. Express your answer using two significant figures. {rm{H}}_2 {rm{O}} . Part D Calculate Kc {rm{N}}_2 . Express your answer using two significant figures. Part C Calculate the equilibrium concentration of H2O {rm{H}}_2 . Express your answer using two significant figures. Part B Calculate the equilibrium concentration of N2

Explanation / Answer

2NO(g) + 2H2(g) <-----> N2(g) + 2H2O (g)

The best strategy for equilibrium problems is to determine initial concentrations [i], any change that occurs (?), and any equilibrium concentration [e] given. Use the change you can calculate to determine the equilibrium concentrations for all species in the reaction. I will use "x" to represent the change in concentration that occurs, using a minus sign on the reactant side, since on that side the concentration decreases, and a plus sign on the product side, since that concentration increases. Coefficients in the balanced equation affect what is lost or gained, so those are combined with x:

.... 2NO + 2H2<>N2 + 2H2O
[i].[.10]....[.050]...[0]..[.10]
?....-2x....-2x....+x....+2x
______________________________
[e][.062]...?.......?.......?

From the info given about NO, 2x = [.038], so x = [.019]

I'm sure you can figure out the equilibrium concentrations yourself now.
[H2] = [.05] - 2(.019)
[N2] = 2(.019)
[H2O] = [.10] + 2(.019)

To calculate K, plug your equilibrium concentrations into the equilibrium expression:
K = [N2][H20]^2/[NO]^2[H2]^2

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