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To predict the effects of an increase in pressure on these chemical reactions, we need to consider Le Chatelier's principle. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves in a way that tends to counteract this change.
### Reaction 1:
[tex]$[tex]$\tag{1} \text{N}_2(g) + 3 \text{H}_2(g) \rightarrow 2 \text{NH}_3(g)$[/tex]$[/tex]
For this reaction, there are 1 mole of \(\text{N}_2\) and 3 moles of \(\text{H}_2\) on the reactant side, making a total of 4 moles of gas. On the product side, there are 2 moles of \(\text{NH}_3\). Increasing the pressure will favor the formation of fewer gas molecules to reduce the pressure. Therefore, the formation of \(2 \text{NH}_3\) (2 moles) is favored.
#### Answer: Favors the formation of \(2 \text{NH}_3(g)\)
### Reaction 2:
[tex]$[tex]$\tag{2} \text{SnO}_2(s) + 2 \text{H}_2(g) \rightarrow \text{Sn}(s) + 2 \text{H}_2\text{O}(g)$[/tex]$[/tex]
In this reaction, there are 2 moles of gas (\(\text{H}_2\)) on the reactant side and 2 moles of gas (\(\text{H}_2\text{O}\)) on the product side. Since the number of moles of gas is the same on both sides, an increase in pressure will not favor either side based solely on the number of gas moles. However, considering the presence of solid-state reactants and products, the decrease in total volume on the reactant side may lead to favoring reactants over products.
#### Answer: Favors the reactants \(\text{SnO}_2(s) + 2 \text{H}_2(g)\)
### Reaction 3:
[tex]$[tex]$\tag{3} \text{AgNO}_3(aq) + \text{Cu}(s) \rightarrow \text{CuNO}_3(aq) + \text{Ag}(s)$[/tex]$[/tex]
This reaction involves no gases; it is entirely in the aqueous and solid phases. Therefore, changes in pressure will have no significant effect on the equilibrium.
#### Answer: No effect, reaction does not involve gases
### Reaction 4:
[tex]$[tex]$\tag{4} 2 \text{SO}_3(g) \rightarrow 2 \text{SO}_2(g) + \text{O}_2(g)$[/tex]$[/tex]
On the reactant side, there are 2 moles of \(\text{SO}_3\), while on the product side, there are 2 moles of \(\text{SO}_2\) and 1 mole of \(\text{O}_2\), making a total of 3 moles of gas. An increase in pressure will favor the side with fewer moles of gas, thus favoring the formation of \(2 \text{SO}_3\) (2 moles).
#### Answer: Favors the formation of \(2 \text{SO}_3(g)\)
### Reaction 5:
[tex]$[tex]$\tag{5} \text{Na}_2\text{CO}_3(s) + 2 \text{HCl}(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l) + 2 \text{NaCl}(aq)$[/tex]$[/tex]
In this reaction, there are no gases on the reactant side and 1 mole of \(\text{CO}_2\) gas on the product side. An increase in pressure will favor the side with fewer moles of gas, which in this case are the reactants (\(\text{Na}_2 \text{CO}_3\) and \(\text{HCl}\)).
#### Answer: Favors the reactants [tex]\(\text{Na}_2\text{CO}_3(s) + 2 \text{HCl}(aq)\)[/tex]
### Reaction 1:
[tex]$[tex]$\tag{1} \text{N}_2(g) + 3 \text{H}_2(g) \rightarrow 2 \text{NH}_3(g)$[/tex]$[/tex]
For this reaction, there are 1 mole of \(\text{N}_2\) and 3 moles of \(\text{H}_2\) on the reactant side, making a total of 4 moles of gas. On the product side, there are 2 moles of \(\text{NH}_3\). Increasing the pressure will favor the formation of fewer gas molecules to reduce the pressure. Therefore, the formation of \(2 \text{NH}_3\) (2 moles) is favored.
#### Answer: Favors the formation of \(2 \text{NH}_3(g)\)
### Reaction 2:
[tex]$[tex]$\tag{2} \text{SnO}_2(s) + 2 \text{H}_2(g) \rightarrow \text{Sn}(s) + 2 \text{H}_2\text{O}(g)$[/tex]$[/tex]
In this reaction, there are 2 moles of gas (\(\text{H}_2\)) on the reactant side and 2 moles of gas (\(\text{H}_2\text{O}\)) on the product side. Since the number of moles of gas is the same on both sides, an increase in pressure will not favor either side based solely on the number of gas moles. However, considering the presence of solid-state reactants and products, the decrease in total volume on the reactant side may lead to favoring reactants over products.
#### Answer: Favors the reactants \(\text{SnO}_2(s) + 2 \text{H}_2(g)\)
### Reaction 3:
[tex]$[tex]$\tag{3} \text{AgNO}_3(aq) + \text{Cu}(s) \rightarrow \text{CuNO}_3(aq) + \text{Ag}(s)$[/tex]$[/tex]
This reaction involves no gases; it is entirely in the aqueous and solid phases. Therefore, changes in pressure will have no significant effect on the equilibrium.
#### Answer: No effect, reaction does not involve gases
### Reaction 4:
[tex]$[tex]$\tag{4} 2 \text{SO}_3(g) \rightarrow 2 \text{SO}_2(g) + \text{O}_2(g)$[/tex]$[/tex]
On the reactant side, there are 2 moles of \(\text{SO}_3\), while on the product side, there are 2 moles of \(\text{SO}_2\) and 1 mole of \(\text{O}_2\), making a total of 3 moles of gas. An increase in pressure will favor the side with fewer moles of gas, thus favoring the formation of \(2 \text{SO}_3\) (2 moles).
#### Answer: Favors the formation of \(2 \text{SO}_3(g)\)
### Reaction 5:
[tex]$[tex]$\tag{5} \text{Na}_2\text{CO}_3(s) + 2 \text{HCl}(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l) + 2 \text{NaCl}(aq)$[/tex]$[/tex]
In this reaction, there are no gases on the reactant side and 1 mole of \(\text{CO}_2\) gas on the product side. An increase in pressure will favor the side with fewer moles of gas, which in this case are the reactants (\(\text{Na}_2 \text{CO}_3\) and \(\text{HCl}\)).
#### Answer: Favors the reactants [tex]\(\text{Na}_2\text{CO}_3(s) + 2 \text{HCl}(aq)\)[/tex]
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