Hydrogen iodide decomposes according to the reaction 2HI(g)leftharpoons H_2(g)+I_2(g) A

The concentration of the products in the equilibrium reaction shown below must be calculated in this problem.
${\displaystyle 2HI\left(g\right)\leftrightarrows H_2\left(g\right)+I_2\left(g\right)}$
Determine the given values first, then solve for each concentration.
${\displaystyle V=15L}$
${\displaystyle {\left[H_2\right]}_i=\frac{0.00623mol}{1.5L}=0.00415M}$
${\displaystyle {\left[I_2\right]}_i=\frac{0.00415mol}{1.5L}=0.00276M}$
${\displaystyle {\left[HI\right]}_i=\frac{0.00244mol}{1.5L}=0.00163M}$
${\displaystyle {\left[H_2\right]}_{eq}=0.0467M}$
Write the reaction table.
$\mid I2HI\leftrightarrows \mid H_2\mid I_2\mid$
$\mid --\mid --\mid --\mid --\mid --\mid$
$\mid I\mid 0.0163\mid \mid 0.00415\mid 0.00276\mid$
$\mid C\mid -2x\mid \mid +x\mid +x\mid$
$\mid E\mid 0.0163-2x\mid \mid 0.00467\mid 0.00276-x\mid$
Since we know the equilibrium concentration of ${\displaystyle H_2}$, we can solve for the change x.
${\displaystyle {\left[H_2\right]}_{eq}={\left[H_2\right]}_i+x}$
${\displaystyle x={\left[H_2\right]}_{eq}-{\left[H_2\right]}_i}$
${\displaystyle =0.00467-0.00415}$
${\displaystyle x=5.2\times {10}^{-4}}$
Now solve for the concentrations of HI and ${\displaystyle I_2}$ using the reaction table and by substituting the calculated value of x.
${\displaystyle {\left[HI\right]}_{eq}={\left[HI\right]}_i-2x}$
${\displaystyle =0.0163-2(5.2\times {10}^{-4})}$
${\displaystyle {\left[HI\right]}_{eq}=0.0153M}$
${\displaystyle {\left[I_2\right]}_{eq}={\left[I_2\right]}_i+x}$
${\displaystyle =0.00276+5.2\times {10}^{-4}}$
${\displaystyle {\left[I_2\right]}_{eq}=0.00328M}$

Given reaction:
${\displaystyle 2HI\left(g\right)\leftrightarrows H_2\left(g\right)+I_2\left(g\right)}$
${\displaystyle H_2:}$
${\displaystyle n=0.00623mol}$
${\displaystyle V=1.5L}$
Firstly we will calculate ${\displaystyle {\left[H_2\right]}_{\in itial}}$ as the ratio of n and V:
${\displaystyle \left[H_2\right]=\frac{n}{V}}$
${\displaystyle \left[H_2\right]=\frac{0.00623mol}{1.5L}}$
${\displaystyle \left[H_2\right]=0.004153M}$
${\displaystyle \left[I_2\right]:}$
${\displaystyle n=0.00414mol}$
${\displaystyle V=1.5L}$
Now we will calculate ${\displaystyle {\left[I_2\right]}_{\in itial}}$ as the ratio of n and V:
${\displaystyle \left[I_2\right]=\frac{n}{V}}$
${\displaystyle \left[I_2\right]=\frac{0.00414mol}{1.5L}}$
${\displaystyle \left[I_2\right]=0.00276M}$
HI:

${\displaystyle n=0.00244mol}$
${\displaystyle V=1.5L}$
Now we will calculate ${\displaystyle {\left[HI\right]}_{\in itial}}$ as the ratio of n and V:
${\displaystyle \left[HI\right]=\frac{n}{V}}$
${\displaystyle \left[HI\right]=\frac{0.0244mol}{1.5L}}$
${\displaystyle \left[HI\right]=0.0163M}$
x - the change in the concentration of HI
Equilibrium concentrations
${\displaystyle HI=0.0163—2x}$
${\displaystyle H_2=0.00415+x}$
${\displaystyle I_2=0.00276+x}$
${\displaystyle {\left[H_2\right]}_{\in itial}=0.00467M}$
Use the equilibrium concentration for ${\displaystyle H_2}$ to find the value of x:
x:
${\displaystyle 0.00415+x=0.00467}$
${\displaystyle x=0.00467—0.00415}$
${\displaystyle x=5.2\cdot {10}^{-4}}$
* So now we can calculate the ${\displaystyle {\left[HI\right]}_{equilibrium}}$ and ${\displaystyle {\left[I_2\right]}_{quiibrium}}$ due we have the value of x:
${\displaystyle \left[HI\right]=0.0163-2x}$
${\displaystyle \left[HI\right]=0.0163-2(5.2\cdot {10}^{-4})}$

Explanation:
First we have to calculate the concentration of $H_2,I_2$ and $HI$
Concentration of $H_2=\frac{Molesof{H}_2}{Volumeofsolution}=\frac{0.00623mol}{1.50L}=0.00415MNSK$
Concentration of $I_2=\frac{Molesof{I}_2}{Volumeofsolution}=\frac{0.00414mol}{1.50L}=0.00276M$
Concentration of $HI=\frac{MolesofHI}{Volumeofsolution}=\frac{0.0244mol}{1.50L}=0.0163M$
Now we have to calculate the value of equilibrium constant (K).
The given chemical reaction is:
$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$
$\begin{array}{}Initialconc.& 0.0163& 0.00415& 0.00276\\ Ateqm.& (0.0163-2x)& (0.00415+x)& (0.00276+x)\end{array}$
As we are given:
Concentration of $H_2$ at equilibrium =0.00467 M
NSK
That means,
(0.00415+x)=0.00467
x=0.00026M
Concentration of HI at equilibrium =(0.0163-2x)=(0.0163-2(0.00026))=0.0158M
Concentration of $I_2$ at equilibrium =(0.00276+x)=(0.00276+0.00026)=0.00302M