Answer:
Step-by-step explanation:
NaNO3
Just do a double displacement reaction, this is what you get (plus water)
The reaction shifts to the left, and the concentrations of SO2(g) and O2(g) increase.
Simply put, the system will seek to stay at equilibrium by shifting to compensate for change.
Complicatedly put, consider the fact that K (the equilibrium constant) is the concentration of products raised to their coefficients over the concentration of the reactants raised to their coefficients. If we write an expression for K, we get this:
![K=([SO_3]^2)/([SO_2]^2[O_2])](https://img.qammunity.org/2022/formulas/chemistry/high-school/2orrpqk5c4tktoxggc22jmikmkritvo2qb.png)
If we assume that, at equilibrium, all of the substances have a concentration of 1 molar, we get that
![K=([1]^2)/([1]^2[1])=1](https://img.qammunity.org/2022/formulas/chemistry/high-school/pa2ce2vqk5f9gahmdek14mepjjgeoarp6h.png)
If we increase SO3, K will increase. This means the system has now been removed from equilibrium. For example, if you add another mole of SO3, you get
![Q=([2]^2)/([1]^2[1])=4](https://img.qammunity.org/2022/formulas/chemistry/high-school/mezkluf0wak4b7md2wlq3v7p8su2pxay6h.png)
(Q is just K, but not at equilibrium.)
Since the system wants to stay at equilibrium, it will work to restore equilibrium. The only way to do this is to increase the value of the reactants (the bottom numbers), meaning the reaction will shift to the left. Worded differently, when Q>K, the reaction will shift left.