Question

For reaction N2
(g) + 3H2(g)  2NH3(g)
6 moles of H2
is taken with excess of N2
gas.
Mass of ammonia gas produced is

Answer 1

Answer:

For reaction N2 + 3H2 = 2NH3. 6 moles of H2 is taken with excess of N2, what is the mass of ammonia gas produced?

If the reaction is considered to be irreversible, then as Azmat has calculated 4 mole of NH3 , i.e., 68 g ( not 68 u) of it will be formed.

But the reaction is reversible, so the amount of NH3 formed will depend on the equilibrium constant which depends on temperature. Further, since the reaction is reversible, one needs to specify the actual amount of 'excess' N2 to get the correct answer. If you apply Le Chatelier's principle , then you may think that the larger the amount of N2 taken, the larger will be the amount of NH3 produced. This is true, however, only if the reaction is carried out at constant volume. At constant pressure, you will find that going on increasing the amount of N2 beyond a certain number of moles will result in smaller amounts of NH3! The reason is simple. At constant pressure adding N2 results in increase in the volume which lowers the concentration of H2. So, although [N2] increases , [H2] decreases so much that [N2]*[H2]^3 also decreases . Since, Kc= [NH3]^2/[N2]*[H2]^3, a decrease in [N2]*[H2]^3 necessitates a decrease in [NH3] to keep Kc constant.

At 623 K, Kp= 7.08*10^-4, and if you take 2.0 mol of N2 and 6.0 mol of H2 at a pressure of 10 atm , you get about 0.55 mol of NH3 with 5.175 mol of H2 and 1.725 mol of N2 remaining unreacted. If, however, you take more than 19.28 mol of N2 to react with 6.0 mol of H2 at the same temperature and pressure you get less than 0.55 mol of NH3. So, taking an excess of N2 at a constant pressure will not change H2 completely into NH3.

At constant volume,at least theoretically, H2 can be completely converted into NH3 by taking an excess of N2. But what is that excess amount? If the reaction is carried out at 623 K in a container of 10L, then for nearly complete conversion of 6.0 mol of H2 to NH3, you require about 6*10^16 mol ofN2 !