Chemistry, 26.02.2020 16:25 michaela134

A mixture of 0.439 M H 2 , 0.317 M I 2 , and 0.877 M HI is enclosed in a vessel and heated to 430 °C. H 2 ( g ) + I 2 ( g ) − ⇀ ↽ − 2 HI ( g ) K c = 54.3 at 430 ∘ C Calculate the equilibrium concentrations of each gas at 430 ∘ C .

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Answer from: Angelofpink2590

Answer : The concentration of $H_2,I_2\text{ and }HI$ at equilibrium is, 0.244 M, 0.122 M and 1.267 M respectively.

Explanation :

The given chemical reaction is:

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

Initial conc. 0.439 0.317 0.877

At eqm. (0.439-x) (0.317-x) (0.877+2x)

As we are given:

K_c=54.3

The expression for equilibrium constant is:

$K=\frac{[HI]^2}{[H_2][I_2]}$

Now put all the given values in this expression, we get:

$54.3=\frac{(0.877+2x)^2}{(0.439-x)\times (0.317-x)}$

x = 0.195 and x = 0.690

We are neglecting the value of x = 0.690 because equilibrium concentration can not be more than initial concentration.

Thus, the value of x = 0.195 M

The concentration of H_2 at equilibrium = (0.439-x) = (0.439-0.195) = 0.244 M

The concentration of I_2 at equilibrium = (0.317-x) = (0.317-0.195) = 0.122 M

The concentration of at equilibrium = (0.877+2x) = (0.877+2\times 0.195) = 1.267 M

Answer from: Quest

You need to divided twenty four by a dozen them by the temperature it is getting cook by

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.

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