Please tell me if these are right.
1.According to Le Chatelier’s principle, what is the effect on the products when N2 is added to the equilibrium mixture of each of the following reactions?
a. 2NH3(g) < = > 3H2(g) + N2(g)
B. N2(g) + O2(g) <= > 2NO(g)
C. 2NO2(g)< => N2(g) +2O2(g)
D. 4NH3(g) + 3O2(g) <= > 2N2(g) +6H2O(g)
A. equilibrium shifts to reactants
B. equilibrium shifts to products
C. equilibrium shifts to reactants
D. equilibrium shifts to products
2.Would increasing the volume of the equilibrium mixture of each of the following reactions cause the equilibrium to shift and if so will the shift be toward products or reactants.
a. 2NH3(g) < => 3H2(g) + N2(g)
B. N2(g) +O2(g) <= > 2NO(g)
C. 2NO2(g) <= > N2(g) + 2O2(g)
D. 4NH3(g) +3O2(g) <= > 2N2(g) + 6H2O(g)
A. equilibrium shifts to reactants
B. equilibrium shifts to products
C. equilibrium shifts to reactants
D. equilibrium shifts to products
1. correct
2. reconsider some of the choices, noting that the volume is increased, so decide which side will be favoured by additional volume.
1.
A. equilibrium shifts to reactants
agreed
B. equilibrium shifts to products
agreed
C. equilibrium shifts to reactants
agreed
D. equilibrium shifts to products
Reactants?
2.
A. equilibrium shifts to reactants
No
B. equilibrium shifts to products
No
C. equilibrium shifts to reactants
No
D. equilibrium shifts to products
agreed
For part 2 if the pressure was increased then the equilibrium shifts in such a way as to oppose the change, i.e. will shift to a smaller volume. As moles are equivalent to volume then you can read these as e.g. equation A
2 mole => 4 moles. So increasing the pressure will move the equilibrium towards the reactants.
In the case of incareasing the volume, the pressure must decrease, so the equilibrium will shift in the opposite direction to a pressure increase. That is to a larger total volume, so the equilibria will shift in the direction that has the larger number of moles.
For equation A
2 mole => 4 moles. So decreasing the pressure or increasing the volume will move the equilibrium towards the products.
1. According to Le Chatelier’s principle, the effect on the products when N2 is added to the equilibrium mixture can be determined as follows:
a. 2NH3(g) <= > 3H2(g) + N2(g)
- Adding N2(g) will increase the concentration of the N2(g) on the right side of the equation.
Therefore, the equilibrium will shift to the left (towards the reactants).
The correct answer is C. equilibrium shifts to reactants.
b. N2(g) + O2(g) <= > 2NO(g)
- Adding N2(g) will increase the concentration of the N2(g) on the left side of the equation.
Therefore, the equilibrium will shift to the right (towards the products).
The correct answer is D. equilibrium shifts to products.
c. 2NO2(g) <= > N2(g) + 2O2(g)
- Adding N2(g) will increase the concentration of the N2(g) on the right side of the equation.
Therefore, the equilibrium will shift to the left (towards the reactants).
The correct answer is C. equilibrium shifts to reactants.
d. 4NH3(g) + 3O2(g) <= > 2N2(g) + 6H2O(g)
- Adding N2(g) will increase the concentration of the N2(g) on the right side of the equation.
Therefore, the equilibrium will shift to the left (towards the reactants).
The correct answer is C. equilibrium shifts to reactants.
2. The effect of increasing the volume of the equilibrium mixture on the shift of equilibrium can be determined as follows:
a. 2NH3(g) <= > 3H2(g) + N2(g)
- Increasing the volume will cause a decrease in pressure.
According to Le Chatelier’s principle, the equilibrium will shift towards the side with more moles of gas to decrease the pressure.
In this case, the right side of the equation has four moles of gas, while the left side has two moles.
Therefore, the equilibrium will shift to the right (towards the products).
The correct answer is B. equilibrium shifts to products.
b. N2(g) + O2(g) <= > 2NO(g)
- Increasing the volume will cause a decrease in pressure.
According to Le Chatelier’s principle, the equilibrium will shift towards the side with more moles of gas to decrease the pressure.
In this case, both sides of the equation have two moles of gas.
Therefore, changing the volume will not cause a shift in equilibrium.
The correct answer is no shift.
c. 2NO2(g) <= > N2(g) + 2O2(g)
- Increasing the volume will cause a decrease in pressure.
According to Le Chatelier’s principle, the equilibrium will shift towards the side with more moles of gas to decrease the pressure.
In this case, both sides of the equation have two moles of gas.
Therefore, changing the volume will not cause a shift in equilibrium.
The correct answer is no shift.
d. 4NH3(g) + 3O2(g) <= > 2N2(g) + 6H2O(g)
- Increasing the volume will cause a decrease in pressure.
According to Le Chatelier’s principle, the equilibrium will shift towards the side with more moles of gas to decrease the pressure.
In this case, the right side of the equation has eight moles of gas, while the left side has seven moles.
Therefore, the equilibrium will shift to the right (towards the products).
The correct answer is D. equilibrium shifts to products.
To determine the effect of adding N2 or increasing the volume on the equilibrium of each reaction, we need to apply Le Chatelier's principle.
1. Effect of adding N2:
According to Le Chatelier's principle, adding N2 to a reaction mixture at equilibrium will cause the equilibrium to shift in the direction that reduces the increase in N2.
For reaction a: 2NH3(g) <=> 3H2(g) + N2(g), adding N2 will shift the equilibrium to the left, towards reactants. So the correct answer is C.
For reaction b: N2(g) + O2(g) <=> 2NO(g), adding N2 will not have any effect on the equilibrium because it already contains N2 in the reactants. So the answer is no shift.
For reaction c: 2NO2(g) <=> N2(g) + 2O2(g), adding N2 will shift the equilibrium to the right, towards products. So the correct answer is D.
For reaction d: 4NH3(g) + 3O2(g) <=> 2N2(g) + 6H2O(g), adding N2 will not have any effect on the equilibrium because there is no N2 present in the reaction. So the answer is no shift.
So the correct answer choices for the effect of adding N2 are: C, no shift, D, no shift.
2. Effect of increasing volume:
When the volume of a reaction mixture at equilibrium is increased, the equilibrium will shift in the direction that decreases the number of gas molecules.
For reaction a: 2NH3(g) <=> 3H2(g) + N2(g), increasing the volume will shift the equilibrium to the side with more moles of gas, which is the product side. So the correct answer is B.
For reaction b: N2(g) + O2(g) <=> 2NO(g), increasing the volume will shift the equilibrium to the side with more moles of gas, which is the product side. So the correct answer is B.
For reaction c: 2NO2(g) <=> N2(g) + 2O2(g), increasing the volume will shift the equilibrium to the side with more moles of gas, which is the reactant side. So the correct answer is A.
For reaction d: 4NH3(g) + 3O2(g) <=> 2N2(g) + 6H2O(g), increasing the volume will shift the equilibrium to the side with more moles of gas, which is the reactant side. So the correct answer is A.
So the correct answer choices for the effect of increasing volume are: B, B, A, A.