Chemistry, 22.05.2020 02:10 Flameking1223

For the reaction

H2(g) + CO2(g) ⇆ H2O(g) + CO(g)

at 700°C, Kc = 0.534. Calculate the number of moles of H2 that are present at equilibrium if a mixture of 0.680 mole of CO and 0.680 mole of H2O is heated to 700°C in a 70.0−L container.

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Answer from: dreamadoreaa

$n_{H_2}^{equilibrium}=0.393mol$

Explanation:

Hello,

In this case, given the amounts of water and carbon dioxide we should invert the given reaction as hydrogen will be producted rather than consumed:

$H_2O(g) + CO(g)\rightleftharpoons H_2(g) + CO_2(g)$

Consequently, the equilibrium constant is also inverted:

$Kc'=\frac{1}{Kc}=\frac{1}{0.534} =1.87$

In such a way, we can now propose the law of mass action:

$Kc'=\frac{[H_2][CO_2]}{[H_2O][CO]}$

And we can express it in terms of the initial concentrations of the reactants and the change due to the reaction extent:

$Kc'=\frac{(x)(x)}{([H_2O]_0-x)([CO]_0-x)}=1.87$

Thus, we compute the initial concentration which are same, since equal amount of moles are given:

$[H_2O]_0=[CO]_0=\frac{0.680mol}{70.0L}=0.0097M$

Hence, solving for by using the quardratic equation or solver, we obtain:

$x_1=0.00561M\\x_2=0.0361M$

For which the correct value is 0.00561M since the other one will produce negative concentrations of water and carbon monoxide at equilibrium. Therefore, the number of moles of hydrogen at equilibrium for the same 70.0-L container turn out:

$n_{H_2}^{equilibrium}=0.00561mol/L*70.0L=0.393mol$

Best regards.

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.

Electrons are transferred from atoms of to atoms of this transfer makes the sodium atoms and the

Answer from: Quest

try answer.com that works for me

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For the reaction

H2(g) + CO2(g) ⇆ H2O(g) + CO(g)

at 700°C, Kc = 0.534. Calcu...