Question

suppose a 250. ml flask is filled with 1.4 mol of i₂ and 0.40 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 formula: $M=\frac{n}{V}$.
Volume $V = 250\ mL=0.25\ L$.
For $I_{2}$: $n = 1.4\ mol$, initial molarity $M_{I_{2}}=\frac{1.4}{0.25}=5.6$.
For $HI$: $n = 0.40\ mol$, initial molarity $M_{HI}=\frac{0.40}{0.25}=1.6$.

Step2: Determine change in molarity based on stoichiometry

From the reaction $H_{2}(g)+I_{2}(g)
ightleftharpoons 2HI(g)$, stoichiometric ratio is $1:1:2$.
If change in $H_{2}$ is $x$, then change in $I_{2}$ is $x$ (because of 1:1 ratio with $H_{2}$), and change in $HI$ is $- 2x$ (because for every 1 mole of $H_{2}$ or $I_{2}$ reacted, 2 moles of $HI$ are consumed or produced. Here, since reaction can go in reverse (as we start with $HI$), if $H_{2}$ is formed, $HI$ is consumed).

Step3: Calculate equilibrium molarity

Equilibrium molarity of $I_{2}$: initial molarity - change. So $5.6 - x$.
Equilibrium molarity of $HI$: initial molarity+change. So $1.6-2x$.

Answer:

	$H_{2}$	$I_{2}$	$HI$
change	$x$	$x$	$-2x$
equilibrium	$x$	$5.6 - x$	$1.6-2x$