Chemistry, 01.08.2019 16:30 afloareiandrei8615

An equilibrium mixture contains 0.750 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00 l container. co(g)+h2o(g)↽−−⇀co2(g)+h2(g) how many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol once equilibrium has been reestablished?

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Answer from: RandomLollipop

1.302 moles of carbon dioxide would have to be added

Further explanation

The equilibrium constant is the value of the product in the equilibrium state of the substance in the right (product) divided by the substance in the left (reactant) with the exponents of each reaction coefficient

The equilibrium constant is based on the concentration (Kc) in a reaction

pA + qB -----> mC + nD

$\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}$

While the equilibrium constant is based on partial pressure

$\large {\boxed {\bold {Kp ~ = ~ \frac {[pC] ^ m [pD] ^ n} {[pA] ^ p [pB] ^ q}}}}$

The value of Kp and Kc can be linked to the formula '

$\large {\boxed {\bold {Kp ~ = ~ Kc. (R.T) ^ {\Delta n}}}}$

R = gas constant = 0.0821 L.atm / mol.K

=n = number of product coefficients-number of reactant coefficients

An equilibrium mixture: 0.750 moles of CO2 and H2, and 0.200 moles of CO and H2O

We determine Kc (constant concentration)

$\displaystyle Kc=\frac{0.75.0.75}{0.2.0.2}\\\\Kc=14.1$

the amount of carbon monoxide to 0.300 mol

Reaction :

CO +H₂O ⇔ CO₂ + H₂

initially 0.2 0.2 0.75+x 0.75

reaction 0.1 0.1 0.1 0.1

product 0.3 0.3 0.65+x 0.65

$\displaystyle Kc=\frac{(0.650+x)(0.65)}{0.3.0.3}\\\\14.1(0.09)=0.4225+0.65x\\\\1.269-0.4225=0.65x\\\\0.8465=0.65x\\\\x=1.302$

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Keywords: constant equilibrium, Kc, concentration, product, reactant, reaction coefficient

Answer from: kaitherusher

Co(g) + H2O(g) ⇄ Co2 (g) + H2 (g)
first, we should get Kc as we have all the concentration of the products and reactant:
when [Co] =0.200 & [H2O] = 0.200 and [Co2] = 0.75 & [H2] = 0.75

∴ Kc= [Co2] [H2] / [Co] [H2O]

= (0.75* 0.75) / (0.200 * 0.200) = 14
then if we want to increase the concentration of Co & H20 to 0.3 mol so we assume the new concenrations are:

[Co] = 0.3 mol& [H2O] = 0.3 mol, so this increase of 0.1 mol in [Co] & [H2O] as well as a 0.1 mol of H2 must react to make this 0.1 of Co and H2O and we assume [CO2] in this case = y - 0.1 & [H2] = 0.75 - 0.1 = 0.65 mol
So we can get the amount of CO2 when:

Kc = [CO2] [H2] / [CO] [ H2O]

14 =( (y-0.1)*(0.65) ) / ( 0.3 * 0.3 ) = 2
∴ y = 2

So to increase the amount of CO to 0.3 the amount of CO2 should increase to 2 mol.

Answer from: Quest

personal, she chose to do it.

Answer from: Quest

pie chart because it is in percents ( % )

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An equilibrium mixture contains 0.750 mol of each of the products (carbon dioxide and hydrogen gas)...

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