Question

The Haber-Bosch process is used to make ammonia from nitrogen and hydrogen: N2(g) + 3 H2(g)  2 NH3(g). A 10.0 L reactor at 450 °C is filled with 3 moles of nitrogen and 12 moles of hydrogen, K is 0.16 under these conditions.

Answer 1

Haber-Bosch Process for manufacture of ammonia

N_{2 (g)} + 3H_{2 (g)}$\leftrightharpoons$ 2NH_{3 (g)}

Equilibrium constant for the preparation of ammonia is given by,

K = $\frac{[NH_{3}]^{^{2}}}{[N_{2}][H_{2}]^{3}}$

where [NH₃], [N₂], [H₂] are molar concentrations of ammonia, nitrogen and hydrogen respectively.

From the chemical reaction, it is found that one mole of nitrogen reacts with three moles of hydrogen and give two moles of ammonia.

[N₂] = 3 mole/10L = 0.3 M

[H₂] = 12 mole/10L = 1.2 M

	N2	H2	NH3
Initial	0.3 M	1.2 M	0 M
change	-x M	-3x M	+2x M
Equilibrium	(0.3-x) M	(1.2-3x) M	2x M

K = 0.16

K = $\frac{[NH_{3}]^{^{2}}}{[N_{2}][H_{2}]^{3}}$

0.16 = $\frac{2x^{^{2}}}{(0.3-x)(1.2-3x)^{3}}$

use the quadratic equation, x = 0.104436 or 1.325705

use 0.104436 to find out concentration of ammonia, nitrogen and hydrogen, reject 1.325705 due to negative concentration terms obtained.

[NH₃] = 2x

= 2 * 0.104436

= 0.208872M

[N₂] = 0.3 - x = 0.3 -0.104436

= 0.195564 M

[H₂] = 1.2 - 3x

= 1.2 -3(0.104436)

= 0.886692 M