Chemistry, 26.07.2019 11:00 pedrojsq271

Amixture of water and graphite is heated to 600 k in a 1 l container. when the system comes to equilibrium it contains 0.15 mol of h2, 0.15 mol of co, 0.57 mol of h2o, and some graphite. some o2 is added to the system and a spark is applied so that the h2 reacts completely with the o2.

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(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???

when the reaction equation is:

C(s) + H2O(g) ↔ H2(g) + CO(g)

∴ Kc = [H2] [CO] / [H2O]

and we have Kc = 0.0393 (given missing in the question)

when the O2 is added so, the reaction will be:

2H2(g) + O2(g) → 2H2O(g)

that means that 0.15 mol H2 gives 0.15 mol of H2O

∴ by using ICE table:

[H2O] [H2] [CO]

initial 0.57 + 0.15 0 0.15

change -X +X +X

Equ (0.72-X) X (0.15+X)

by substitution:

0.0393 = X (0.15+X) / (0.72-X) by solving for X

∴ X = 0.098

∴[H2] = X = 0.098 M

∴[CO] = 0.15 + X

= 0.15 + 0.098 = 0.248 M

∴[H2O] = 0.72 - X

= 0.72 - 0.098

= 0.622 M

Answer from: heroxtree

The concentration of $\rm H_2$ is 0.098 mol, CO is 0.248 mol, and $\rm H_2O$ is 0.622 mol after equilibrium.

The complete question is to find the concentration of CO, $\rm H_2$ and $\rm H_2O$ . Given, $\rm k_c$ = 0.0393.

The reaction will be:

$\rm H_2O\;+\;C\;\rightarrow\;H_2\;+CO$

By the addition of excess $\rm O_2$ ,

$\rm H_2\;+O_2\;\rightarrow\;H_2O$

At equillibrium,

$\rm H_2$ = 0.15 mol

CO= 0.15 mol

$\rm H_2O$ = 0.57 mol

Initially, $\rm H_2O$ = 0.57 + 0.15

CO = 0.15

After equillibrium,

$\rm H_2O$ = 0.57 + 0.15 - X

$\rm CO$ = 0.15 + X

$\rm H_2$ = +X

0.0393 = $\rm \frac{0.15\;+\;X}{0.72\;-X}$

X = 0.098

Concentration after equillibrium :

$\rm H_2$ = 0.098 mol

CO= 0.15 + 0.098 = 0.248 mol

$\rm H_2O$ = 0.72 - 0.098 = 0.622 mol

For more information about equilibrium, refer the link:

link

Answer from: Quest

possibly, yes.

explanation:

the amount of sugar you take in is dangerously high, risk of heart disease, high risk of diabetes, and etc.

Answer from: Quest

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^{1}$ and electronic configuration of phosphorous is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

$3na + p \rightarrow na_{3}p$

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.