Question

nitrogen and hydrogen react to form ammonia, like this:
n₂(g)+3h₂(g)→2nh₃(g)
use this chemical equation to answer the questions in the table below.
suppose 140. mmol of n₂ and 420. mmol of h₂ are added to an empty flask. how much nh₃ will be in the flask at equilibrium?
suppose 215. mmol of nh₃ are added to an empty flask. how much n₂ will be in the flask at equilibrium?

Explanation:

First Question:

Step1: Analyze the stoichiometry

From the equation \(N_{2}(g)+3H_{2}(g)\to2NH_{3}(g)\), the mole ratio of \(N_{2}:H_{2}:NH_{3}\) is \(1:3:2\). If there was no equilibrium (complete reaction), \(140\) mmol of \(N_{2}\) reacting with \(420\) mmol of \(H_{2}\) (\(140\times3 = 420\)) would produce \(n(NH_{3})=2\times140=280\) mmol. But in a reversible reaction (equilibrium), the reaction does not go to completion.

Second Question:

Step1: Consider the reverse reaction

The reverse reaction is \(2NH_{3}(g)
ightleftharpoons N_{2}(g)+3H_{2}(g)\). If there was no equilibrium (complete decomposition), \(215\) mmol of \(NH_{3}\) would produce \(n(N_{2})=\frac{215}{2}=107.5\) mmol. But since it is an equilibrium reaction, \(NH_{3}\) does not completely decompose.

Answer:

• First question: Some, but less than \(280\). mmol.
• Second question: Some, but less than \(108\). mmol.