Consider the following reaction: N2(g) + 3H2(g) 2NH3(g) In a given experiment, 1

Question

Consider the following reaction: N2(g) + 3H2(g) 2NH3(g)

In a given experiment, 1.00 moles of N2(g) and 4.01 moles of H2(g) were placed in a 4.63 L vessel. Complete the following ICE 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 0.231 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 =

Explanation / Answer

1mole/4.63 L= 0.216 M

4.01moles/4.63 L = 0.866 M

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

Initial 0.216 0.866 0

Change -x -3x +2x

Equilibrium 0.216-x 0.866-3x 2x

K = (2x)2/(0.216-x)(0.866-3x)3   

If 2x = 0.231 M

The x = 0.231/2 = 0.1155 M

So N2 = 0.216 - 0.1155 = 0.1005 M

H2 = 0.866 - 3 x 0.1155 = 0.5195 M

K = 0.2312/0.1005 x 0.51953

K =3.79

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