Consider the following reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) In a given experim

ID: 881569 • Letter: C

Question

Consider the following reaction:

N2(g) + 3H2(g) 2NH3(g)

In a given experiment, 1.21 moles of N2(g) and 4.52 moles of H2(g) were placed in a 5.65 L vessel. Complete the following table by entering numerical values in the Initial row and values containing the variable "x" in the Change and Equilibrium rows. Define x as the amount (mol/L) of N2 that reacts to reach equilibrium. Include signs in the Change column to indicate a gain or loss of concentration. (Omit units and spaces, use 3 sig. fig. and write concentration less than 1 as 0.###, not as .###. If nothing is present initially, enter 0 for the molarity.)

At a particular temperature, the equilibrium concentration of NH3 is found to be 0.229 M.

Given this information, determine the numerical value of K for this reaction, at this temperature.
(If you want to enter your answer in scientific notation, use the format as shown below:
2.5 x 10-3 = 2.5E-3)

K = ?

[N2] (M) [H2] (M) [NH3] (M) Initial 0.214 0.800 0 Change -x -3x +2x Equilibrium 0.214-x 0.800-3x 2x

Explanation / Answer

Answer –

Given, N2 = 1.21 mole , H2 4.52 moles , volume = 5.65 L

[N2] (M)

[H2] (M)

[NH3] (M)

Initial

0.214

0.800

Change

-x

-3x

+2x

Equilibrium

0.214-x

0.800-3x

We also given, at the equilibrium [NH3] = 0.229 M

We know at the equilibrium, [NH3] = 2x = 0.229 M

2x = 0.229

So, x = 0.229 / 2

= 0.1145 M

[N2] = 0.214-x

= 0.214-0.1145

= 0.0995 M

[H2] = 0.800-3x

= 0.800 -3*0.1145 M

= 0.4565 M

We know,

K = [NH3]2 / [N2] [H2]3

= (0.229)2 / (0.0995 ) * (0.4565)3

= 5.54