Question

Consider the following reaction: N2(9) + 3H2(9) = 2NH3(9) In a given experiment, 1.42 moles of N (9) and 4.01 moles of H2(g) were placed in a 4.12 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 Ny 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 o for the molarity.) [H2] (M) (NH3] (M) [N2] (M) 0.345 0.995 Initial Change Equilibrium Submit Answer Incorrect. Tries 14/100 Previous Tries At a particular temperature, the equilibrium concentration of NH3 is found to be 0.368 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=

Answer 1

Part A

N₂ + 3 H₂$\small \rightleftharpoons$ 2 NH₃

ICE table is

[N₂] = moles/Volume = (1.42/4.12) = 0.345 M

[H₂] = (4.01/4.12) =  0.973 M

	[N2] (M)	[H2] (M)	[NH3] (M)
I	0.345	0.973	0
C	- x	-3x	+2x
E	0.345 - x	0.973 - 3x	2x

Part B

at equilibrium [NH₃] = 0.368 M

then the ICE table becomes

	[N2] (M)	[H2] (M)	[NH3] (M)
I	0.345	0.973	0
C	- (0.368/2) = 0.184	- 3$\small \times$0.184 = - 0.552	+ 0.368
E	0.345 - 0.184 = 0.161	(0.973 - 0.552 ) = 0.421	0.368

Equilibrium constant

K =  $\small \frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}$ =  $\small \frac{(0.368^{2})}{0.161\times (0.421)^{3}}$= 11.3 = 1.13 $\small \times$10¹