Question

suppose a 250. ml flask is filled with 1.2 mol of h₂ and 0.70 mol of hi. this reaction becomes possible:
h₂(g) + i₂(g) ⇌ 2hi(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 h₂. you can leave out the m symbol for molarity.

Explanation:

Step1: Calculate initial molarity

Molarity \(M=\frac{n}{V}\) (where \(n\) is moles and \(V\) is volume in liters). Given \(V = 250\space mL=0.25\space L\).
For \(H_2\): \(M_{H_2}=\frac{1.2}{0.25}=4.8\)
For \(I_2\): Initially, \(n = 0\), so \(M_{I_2}=0\)
For \(HI\): \(M_{HI}=\frac{0.70}{0.25}=2.8\)

Step2: Determine change in molarity based on stoichiometry

The balanced equation is \(H_2(g)+I_2(g)
ightleftharpoons2HI(g)\).
If the change in molarity of \(H_2\) is \(x\) (since \(x\) is defined as the change in \(H_2\) molarity), then for \(I_2\) (stoichiometric ratio \(1:1\) with \(H_2\)), the change is \(x\). For \(HI\) (stoichiometric ratio \(2:1\) with \(H_2\)), the change is \(- 2x\) (because it's a reverse - like consideration for the change as we are starting with some \(HI\) already present. If we assume the reaction can shift in either direction, based on the stoichiometry of the balanced equation).

Step3: Calculate equilibrium molarity

Equilibrium molarity of \(H_2\): \(M_{H_2}(eq)=4.8 + x\) (if we consider the general form of equilibrium molarity \(=\) initial molarity \(+\) change. Here, if \(x\) is negative, it means \(H_2\) is being consumed).
Equilibrium molarity of \(I_2\): \(M_{I_2}(eq)=0+x\)
Equilibrium molarity of \(HI\): \(M_{HI}(eq)=2.8-2x\)

Answer:

	\(H_2\)	\(I_2\)	\(HI\)
Change	\(x\)	\(x\)	\(-2x\)
Equilibrium	\(4.8 + x\)	\(x\)	\(2.8-2x\)