Answer:
The equilibrium value of [CO] is 1.04 M
Explanation:
Chemical equilibrium is the state to which a spontaneously evolving chemical system, in which a reversible chemical reaction takes place. When this situation is reached, it is observed that the concentrations of substances, both reagents and reaction products, they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.
Reagent concentrations and products in equilibrium are related by the equilibrium constant Kc. Being:
aA + bB ⇔ cC + dD
![Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%20%2A%5BB%5D%5E%7Bb%7D%20%7D)
Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.
In this case:
![Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BCH_%7B3%7DOH%20%5D%7D%7B%5BCO%5D%2A%5BH_%7B2%7D%20%5D%5E%7B2%7D%20%7D)
You know:
- Kc= 14.5
- [H₂]= 0.322 M
- [CH₃OH] =1.56 M
Replacing:
![14.5=\frac{1.56}{[CO]*0.322^{2} }](https://tex.z-dn.net/?f=14.5%3D%5Cfrac%7B1.56%7D%7B%5BCO%5D%2A0.322%5E%7B2%7D%20%7D)
Solving:
![[CO]=\frac{1.56}{14.5*0.322^{2} }](https://tex.z-dn.net/?f=%5BCO%5D%3D%5Cfrac%7B1.56%7D%7B14.5%2A0.322%5E%7B2%7D%20%7D)
[CO]= 1.04 M
The equilibrium value of [CO] is 1.04 M
Answer : The molarity after a reaction time of 5.00 days is, 0.109 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken = 5.00 days
[A] = concentration of substance after time 't' = ?
= Initial concentration = 0.110 M
Now put all the given values in above equation, we get:
![9.7\times 10^{-6}=\frac{1}{5.00}\left (\frac{1}{[A]}-\frac{1}{(0.110)}\right)](https://tex.z-dn.net/?f=9.7%5Ctimes%2010%5E%7B-6%7D%3D%5Cfrac%7B1%7D%7B5.00%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.110%29%7D%5Cright%29)
![[A]=0.109M](https://tex.z-dn.net/?f=%5BA%5D%3D0.109M)
Hence, the molarity after a reaction time of 5.00 days is, 0.109 M
The object and description that matches is Object 2 and T.
Object 1 has no matching description.
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Therefore, the object and description that matches is Object 2 and T.
Object 1 has no matching description.
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Answer: D. 19.9 g hydrogen remains.
Explanation:
To calculate the moles, we use the equation:
a) moles of
b) moles of
According to stoichiometry :
1 mole of
require 1 mole of
Thus 0.0787 moles of
require=
of
Thus
is the limiting reagent as it limits the formation of product and
acts as the excess reagent. (10.0-0.0787)= 9.92 moles of
are left unreacted.
Mass of
Thus 19.9 g of
remains unreacted.