Frank Guyton

2022-01-06

Hydrogen iodide decomposes according to the reaction

Nadine Salcido

Beginner2022-01-07Added 34 answers

The concentration of the products in the equilibrium reaction shown below must be calculated in this problem.

$2HI\left(g\right)\leftrightarrows {H}_{2}\left(g\right)+{I}_{2}\left(g\right)$

Determine the given values first, then solve for each concentration.

$V=15L$

${\left[{H}_{2}\right]}_{i}=\frac{0.00623mol}{1.5L}=0.00415M$

${\left[{I}_{2}\right]}_{i}=\frac{0.00415mol}{1.5L}=0.00276M$

${\left[HI\right]}_{i}=\frac{0.00244mol}{1.5L}=0.00163M$

${\left[{H}_{2}\right]}_{eq}=0.0467M$

Write the reaction table.

$\mid I2HI\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 $H}_{2$, we can solve for the change x.

${\left[{H}_{2}\right]}_{eq}={\left[{H}_{2}\right]}_{i}+x$

$x={\left[{H}_{2}\right]}_{eq}-{\left[{H}_{2}\right]}_{i}$

$=0.00467-0.00415$

$x=5.2\times {10}^{-4}$

Now solve for the concentrations of HI and $I}_{2$ using the reaction table and by substituting the calculated value of x.

${\left[HI\right]}_{eq}={\left[HI\right]}_{i}-2x$

$=0.0163-2(5.2\times {10}^{-4})$

${\left[HI\right]}_{eq}=0.0153M$

${\left[{I}_{2}\right]}_{eq}={\left[{I}_{2}\right]}_{i}+x$

$=0.00276+5.2\times {10}^{-4}$

${\left[{I}_{2}\right]}_{eq}=0.00328M$

Bubich13

Beginner2022-01-08Added 36 answers

Given reaction:

Firstly we will calculate

Now we will calculate

HI:

Now we will calculate

x - the change in the concentration of HI

Equilibrium concentrations

Use the equilibrium concentration for

x:

* So now we can calculate the

karton

Expert2022-01-10Added 613 answers

Explanation:

First we have to calculate the concentration of

Concentration of

Concentration of

Concentration of

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

As we are given:

Concentration of

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

22+64

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