Frank Guyton

2022-01-06

Hydrogen iodide decomposes according to the reaction $2HI\left(g\right)⇆{H}_{2}\left(g\right)+{I}_{2}\left(g\right)$ A sealed $1.50-L$ container initially holds 0.00623 mol of H2, 0.00414 mol of 12, and 0.0244 mol of H1 at 703 K. When equilibrium is reached, the concentration of H2(g) is 0.00467 M. What are the concentrations of H1(g) and 12(g)?

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

Bubich13

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

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

karton

Explanation:
First we have to calculate the concentration of ${H}_{2},{I}_{2}$ and $HI$
Concentration of
Concentration of
Concentration of
Now we have to calculate the value of equilibrium constant (K).
The given chemical reaction is:
$2HI\left(g\right)⇌{H}_{2}\left(g\right)+{I}_{2}\left(g\right)$
$\begin{array}{}Initialconc.& 0.0163& 0.00415& 0.00276\\ Ateqm.& \left(0.0163-2x\right)& \left(0.00415+x\right)& \left(0.00276+x\right)\end{array}$
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

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