Question

suppose a 500. ml flask is filled with 1.6 mol of n₂ and 0.60 mol of nh₃. this reaction becomes possible:
n₂(g)+3h₂(g)⇌2nh₃(g)
complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium.
use x to stand for the unknown change in the molarity of n₂. you can leave out the m symbol for molarity.

Explanation:

Step1: Calculate initial molarity

Molarity \(c=\frac{n}{V}\). For \(N_2\), \(n = 1.6\space mol\), \(V=0.500\space L\), so initial \(c(N_2)=\frac{1.6}{0.500}=3.2\). For \(NH_3\), \(n = 0.60\space mol\), \(V = 0.500\space L\), so initial \(c(NH_3)=\frac{0.60}{0.500}=1.2\). Initially, \(H_2\) is not added, so initial \(c(H_2)=0\).

Step2: Determine change in molarity

From the reaction \(N_2(g)+3H_2(g)
ightleftharpoons 2NH_3(g)\), if change in \(N_2\) is \(x\) (since \(x\) is defined as change in \(N_2\) molarity), then change in \(H_2\) is \(3x\) (because of the stoichiometric coefficient 3 for \(H_2\) in the reaction) and change in \(NH_3\) is \(- 2x\) (because \(NH_3\) is being consumed if the reaction shifts in the reverse direction as \(N_2\) is increasing).

Step3: Calculate equilibrium molarity

Equilibrium molarity of \(N_2\) is initial \(+\) change, so \(3.2 + x\). Equilibrium molarity of \(H_2\) is initial \(+\) change, so \(0+3x = 3x\). Equilibrium molarity of \(NH_3\) is initial \(+\) change, so \(1.2-2x\).

Answer:

	\(N_2\)	\(H_2\)	\(NH_3\)
Change	\(x\)	\(3x\)	\(-2x\)
Equilibrium	\(3.2 + x\)	\(3x\)	\(1.2-2x\)