Answer:
The system is not in equilibrium and will evolve left to right to reach equilibrium.
Explanation:
The reaction quotient Qc is defined for a generic reaction:
aA + bB → cC + dD
![Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%2A%5BB%5D%5E%7Bb%7D%20%20%7D)
where the concentrations are not those of equilibrium, but other given concentrations
Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:
![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%2A%5BB%5D%5E%7Bb%7D%20%20%7D)
where the concentrations are those of equilibrium.
This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.
Comparing Qc with Kc allows to find out the status and evolution of the system:
- If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.
- If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.
- If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.
In this case:
![Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BSo_%7B3%7D%5D%20%5E%7B2%7D%20%7D%7B%5BSO_%7B2%7D%20%5D%5E%7B2%7D%2A%20%5BO_%7B2%7D%5D%20%7D)

Q=100,000
100,000 < 4,300,000 (4.3*10⁶)
Q < Kc
<u><em>
The system is not in equilibrium and will evolve left to right to reach equilibrium.</em></u>
The equilibrium constant expression for KSP of Sr3(PO4)2 is
KSP={(Sr^2+)^3 (PO4^3-)^2/ Sr3(PO4)2}
Explanation
write the ionic equation for Sr3(PO4)2
Sr3(PO4)2 → 3Sr^2+ + 2 PO4^3-
KSP is given by (concentration of the products raised to their coefficient /concentration of reactants raised to their coefficient)
Sulfur reacts with oxygen to produce sulfur dioxide. That is for every mole of sulfur reacted, one mole of sulfur dioxide also is produced. With the given mole of sulfur dioxide, the amount of sulfur in mass is determined by multiplying the number of moles to the molar mass of sulfur (32 g/mol).
"One has more oxygen atoms than the other" best distinguishes carbon dioxide from carbon monoxide
<u>Explanation</u>:
In both carbon monoxide and carbon dioxide, there is the presence of carbon and oxygen. However,they vary depending on the number of oxygen atoms.In carbon monoxide, there is the presence of one only carbon and oxygen atom. Thus carbon monoxide is chemically represented as CO. Only about 0.2%ppm of carbon monoxide present in air. carbon monoxide is toxic and causes death in human when not consumed is correct proportion.
Whereas in carbon dioxide, only one carbon atom,and two oxygen atom and hence chemically represented as co2.