In a given experiment 1.00 mole of N2 and 3.00 moles of H2 were placed in a 3.00

Question

In a given experiment 1.00 mole of N2 and 3.00 moles of H2 were placed in a 3.00 L container. Equilibrium was established given the following reaction: N2(g) + 3H2(g) 2NH3(g) Complete the following table. Use numerical values in the Initial row and values containing the variable "x" in both the Change and Equilibrium rows. Let x = the amount of N2 needed to reach equilibrium.

At equilibrium the concentration of NH3 is 0.0519 M.

Determine the following quantities at equilibrium. Calculate the concentrations to 2 significant figures:

[N2] = M
[H2] = M

Determine the numerical value of K for this reaction.

(Enter your answer to two significant figures.)

K =

Explanation / Answer

               N2(g)      +             3H2(g) <---> 2NH3(g)

Initial:    1 mole                  3 moles                0

Eqm:      1-x                        3-3x                      2x

concentration of NH3 at eqm = 0.0519 M

So amount of NH3 at eqm = 0.0519 M * 3 L = 0.1557 moles

2x = 0.1557 moles

x = 0.07785 moles

amount of N2 at eqm = 1-x = 1- 0.07785 = 0.92215 moles

concentration of N2 at eqm = 0.92215 moles/ 3 L = 0.31 M

amount of H2 at eqm = 3-3x = 3- (3*0.07785) = 2.76645 moles

concentration of H2 at eqm = 2.76645 moles/ 3 L = 0.92 M

Keqm = [NH3]2/[ N2][ H2]3 = (0.0519)2/(0.31)(0.92)3 = 0.01

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