How many NESA Chemistry questions cover Factors that Affect Equilibrium?

AusGrader has 99 NESA Chemistry questions on Factors that Affect Equilibrium, all with instant AI grading and detailed marking feedback.

NESA Chemistry Factors that Affect Equilibrium

15 sample questions with marking guides and sample answers · Avg. score: 63.1%

Q38

2021

SCSA

12 marks

Q38

12 marks

Sulfuric acid is manufactured by the Contact process, the steps of which are outlined below.

Step One: Molten sulfur is burned in air at approximately 1000 °C:

S(l) + O_2(g) \rightarrow SO_2(g) + 297 kJ

Step Two: The resulting sulfur dioxide is converted to sulfur trioxide as shown in the following equilibrium reaction. It is conducted at a temperature of about 450 °C with a $V_2O_5$ catalyst at a pressure of between 100 and 200 kPa:

2 SO_2(g) + O_2(g) \rightleftharpoons 2 SO_3(g) + 198 kJ

Step Three: The resulting sulfur trioxide is absorbed into sulfuric acid, producing oleum ( $H_2S_2O_7$ ). Water is added to the oleum, producing 18 mol L $^{-1}$ sulfuric acid:

SO_3(g) + H_2SO_4(l) \rightarrow H_2S_2O_7(l)

H_2S_2O_7(l) + H_2O(l) \rightarrow 2 H_2SO_4(aq)

Use your understanding of collision theory and chemical equilibrium to discuss the reaction conditions for Steps 1 and 2 of the Contact process, given that the aim is to produce the greatest yield in the shortest time. In your discussion, also address economic concerns where appropriate.

Reveal Answer

High temperature increases the average kinetic energy of the particles, which means that the particles collide more frequently. Also, more of these collisions will have energy higher than the activation energy, which means a greater proportion of collisions are successful, and the reaction rate increases. The vanadium catalyst increases the rate of the forward reaction, and also the rate of the reverse reaction to an equal extent, as it provides an alternative pathway with a lower activation energy. Therefore, a greater proportion of the particles will have sufficient energy to react when they collide. High pressure or concentration has more particles per unit volume and so there is a higher frequency of collisions, and the reaction rate increases. As Step 1 is a combustion reaction, it essentially goes to completion at the high temperature and does not require a catalyst or high pressure. For Step 2, high temperature, high pressure and a catalyst would favour a high rate.

For equilibrium, which is only considered for Step 2, high temperature favours the reverse reaction because it is endothermic, and this decreases the SO $_3$ (g) yield, which is not desired. A low temperature decreases the rate of reaction, which is also not desired. A high pressure favours the forward reaction because there are a greater number of moles of gas reactants, increasing the SO $_3$ (g) yield which is desired.

Economically, high pressures are costly and dangerous.

Therefore, for Step 2, a compromise is required between the high temperature for rate and the low temperature for yield. A compromise is also required between the cost of higher pressures and the pressure that allows a satisfactory yield and rate.

Marking Criteria

Rates

Descriptor	Marks
1 mark each for any of the following, up to a maximum of 6 marks: Explains that high temperature increases the average kinetic energy of the particles, which means that the particles collide more frequently; Explains that more of these collisions will have energy higher than the activation energy, which means a greater proportion of collisions are successful, and the reaction rate increases; Describes that the vanadium catalyst increases the rate of the forward reaction (and also the rate of the reverse reaction to an equal extent) as it provides an alternative pathway with a lower activation energy; States that a greater proportion of the particles will have sufficient energy to react when they collide; Explains that high pressure (concentration) has more particles per unit volume and so there is a higher frequency of collisions, and the reaction rate increases; Identifies that as Step 1 is a combustion reaction, it essentially goes to completion at the high temperature (and does not require a catalyst or high pressure); States that for Step 2, high temperature, high pressure and catalyst would favour high rate.	6

Descriptor

Marks

1 mark each for any of the following, up to a maximum of 6 marks: Explains that high temperature increases the average kinetic energy of the particles, which means that the particles collide more frequently; Explains that more of these collisions will have energy higher than the activation energy, which means a greater proportion of collisions are successful, and the reaction rate increases; Describes that the vanadium catalyst increases the rate of the forward reaction (and also the rate of the reverse reaction to an equal extent) as it provides an alternative pathway with a lower activation energy; States that a greater proportion of the particles will have sufficient energy to react when they collide; Explains that high pressure (concentration) has more particles per unit volume and so there is a higher frequency of collisions, and the reaction rate increases; Identifies that as Step 1 is a combustion reaction, it essentially goes to completion at the high temperature (and does not require a catalyst or high pressure); States that for Step 2, high temperature, high pressure and catalyst would favour high rate.

Equilibrium

Descriptor	Marks
1 mark each for any of the following (only considered for Step 2), up to a maximum of 3 marks: Explains that high temperature favours the reverse reaction because it is endothermic, and this decreases the SO3(g) yield; States that a low temperature decreases the rate of reaction; Explains that a high pressure favours the forward reaction because there are a greater number of moles of gas reactants, increasing the SO3(g) yield.	3

Descriptor

Marks

1 mark each for any of the following (only considered for Step 2), up to a maximum of 3 marks: Explains that high temperature favours the reverse reaction because it is endothermic, and this decreases the SO3(g) yield; States that a low temperature decreases the rate of reaction; Explains that a high pressure favours the forward reaction because there are a greater number of moles of gas reactants, increasing the SO3(g) yield.

Economics

Descriptor	Marks
States that high pressures are costly (and dangerous).	1

Compromise

Descriptor	Marks
1 mark each for any of the following, up to a maximum of 2 marks: Identifies that for Step 2, a compromise is required between the high temperature for rate and the low temperature for yield; Identifies that a compromise is also required between the cost of higher pressures and the pressure that allows a satisfactory yield and rate.	2

2024

QCAA

Paper 2

2 marks

In the aqueous solution of a chromate salt, an equilibrium exists between the yellow chromate ( $\text{CrO}_4^{2-}$ ) ions and the orange dichromate ( $\text{Cr}_2\text{O}_7^{2-}$ ) ions. This equilibrium can be represented by the equation shown.

$2\text{CrO}_4^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) \rightleftharpoons \text{Cr}_2\text{O}_7^{2-}(\text{aq}) + \text{H}_2\text{O}(\text{l})$
(yellow) $\hspace{3cm}$ (orange)

Explain, at an atomic level, why no colour change occurs once the chromate–dichromate solution has established equilibrium.

Reveal Answer

At equilibrium, the forward reaction rate equals the reverse reaction rate. Therefore, yellow chromate ions and orange dichromate ions are re-formed at the same rate at which they are broken down, so the colour remains constant because the $[CrO_4^{2-}]$ and $[Cr_2O_7^{2-}]$ remain constant.

Marking Criteria

Descriptor	Marks
Identifies that the forward reaction rate equals the reverse reaction rate at equilibrium	1
Explains that the colour remains constant because the $[CrO_4^{2-}]$ and $[Cr_2O_7^{2-}]$ remain constant	1

Q24

2021

VCAA

1 mark

Q24

1 mark

Which one of the following statements describes the effect that adding a catalyst will have on the energy profile diagram for an exothermic reaction?

The energy of the products will remain the same.

The shape of the energy profile diagram will remain the same.

The peak of the energy profile will move to the left as the reaction rate increases.

The activation energy will be lowered by the same proportion in the forward and reverse reactions.

Reveal Answer

The energy of the products will remain the same.

Correct Answer

A catalyst provides an alternative reaction pathway with a lower activation energy, but it does not alter the initial energy of the reactants or the final energy of the products.

The shape of the energy profile diagram will remain the same.

The shape of the energy profile diagram changes because the peak, which represents the activation energy, is lowered when a catalyst is added.

The peak of the energy profile will move to the left as the reaction rate increases.

The x-axis of an energy profile diagram represents reaction progress, not time. A catalyst lowers the peak vertically rather than shifting it horizontally.

The activation energy will be lowered by the same proportion in the forward and reverse reactions.

A catalyst lowers the activation energy of both the forward and reverse reactions by the same absolute amount, not the same proportion, since their initial activation energies are different.

2025

SCSA

1 mark

The equation for a system at equilibrium is given below.

$2 NO(g) + O_2(g) \leftrightharpoons 2 NO_2(g) + heat$

At 25 `C, the value of K for this equilibrium is $2.19 \times 10^{12}$ .

Which of the following statements about this system is true? Increasing the

partial pressure of NO(g) will increase the yield of $NO_2(g)$ and will increase the rate of the forward reaction.

partial pressure of NO(g) will increase the yield of $NO_2(g)$ but will decrease the rate of the forward reaction.

temperature will increase the yield of $NO_2(g)$ but decrease the rate of the forward reaction.

temperature will increase the yield of $NO_2(g)$ and increase the rate of the forward reaction.

Reveal Answer

partial pressure of NO(g) will increase the yield of $NO_2(g)$ and will increase the rate of the forward reaction.

Correct Answer

According to Le Chatelier's principle, increasing the partial pressure of a reactant ( $NO$ ) shifts the equilibrium toward the products, increasing the yield. Additionally, a higher concentration of reactants increases the frequency of collisions, thereby increasing the forward reaction rate.

partial pressure of NO(g) will increase the yield of $NO_2(g)$ but will decrease the rate of the forward reaction.

While increasing the partial pressure of a reactant does increase the yield, it also increases (rather than decreases) the reaction rate due to a higher frequency of molecular collisions.

temperature will increase the yield of $NO_2(g)$ but decrease the rate of the forward reaction.

Increasing temperature increases the kinetic energy of the molecules, which always leads to an increase in the reaction rate, not a decrease.

temperature will increase the yield of $NO_2(g)$ and increase the rate of the forward reaction.

The reaction forming $NO_2$ is exothermic; therefore, increasing the temperature shifts the equilibrium toward the reactants (left), decreasing the yield of $NO_2$ .

Q21

2023

QCAA

Paper 1

4 marks

Q21

4 marks

$\text{CO(g)}$ reacts with $\text{O}_2\text{(g)}$ in a sealed container producing $\text{CO}_2\text{(g)}$ to reach equilibrium.

$2\text{CO(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{CO}_2\text{(g)}$

Apply collision theory to explain how increasing the concentration of $\text{O}_2$ at equilibrium will affect the concentration of $\text{CO}_2$ if the temperature and volume are held constant.

Reveal Answer

Increasing the concentration of $O_2$ increases the number of $O_2$ molecules.
This increases the frequency of collisions between $O_2$ and $CO$ .
Therefore, the rate of the forward reaction increases and equilibrium shifts to the right (products) to increase the concentration of $CO_2$ .

Marking Criteria

Descriptor	Marks
identifies that increasing concentration of $O_2$ increases the number of $O_2$ molecules	1
explains that an increase in $[O_2]$ increases collisions between $O_2$ and $CO$	1
explains that the rate of forward reaction increases	1
identifies that equilibrium shifts to the right and concentration of $CO_2$ increases	1

Q12

2023

VCAA

1 mark

Q12

1 mark

Consider the following features of a chemical reaction.

I. activation energy
II. $\Delta H$ of the reaction
III. enthalpy of the reactants

Compared with the uncatalysed reaction pathway, the presence of a catalyst changes

I only.

II only.

II and III only.

I, II and III.

Reveal Answer

I only.

Correct Answer

A catalyst provides an alternative reaction pathway with a lower activation energy, but it does not affect the enthalpy of the reactants or the overall $\Delta H$ of the reaction.

II only.

A catalyst does not change the overall enthalpy change ( $\Delta H$ ) of the reaction, as the initial and final states of the system remain the same.

II and III only.

Neither the $\Delta H$ of the reaction nor the enthalpy of the reactants are changed by a catalyst; only the activation energy is affected.

I, II and III.

While a catalyst does change the activation energy, it does not alter the enthalpy of the reactants or the overall $\Delta H$ of the reaction.

Q25

2021

QCAA

Paper 1

5 marks

Q25

An equilibrium is formed between two differently coloured cobalt species, $\text{Co(H}_2\text{O})_6^{2+}$ (aq), which is pink, and $\text{CoCl}_4^{2-}$ (aq), which is blue. The equation for this equilibrium is shown.

$\text{Co(H}_2\text{O})_6^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightleftharpoons \text{CoCl}_4^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})$

Q25a

3 marks

Apply Le Châtelier’s principle to predict the visible effect of adding $\text{AgNO}_3$ to an aqueous blue-coloured solution containing $\text{Co(H}_2\text{O})_6^{2+}$ and $\text{CoCl}_4^{2-}$ ions.
Explain your reasoning.

Reveal Answer

Adding $AgNO_3$ produces $Ag^+$ ions, which react with $Cl^-$ ions to form insoluble AgCl, therefore decreasing the concentration of $Cl^-$ ions.
Equilibrium shifts to reactants (left) to counteract the decrease by increasing the concentration of $Cl^-$ ions.
The blue solution will become lighter (pinker).

Marking Criteria

Descriptor	Marks
Correctly identifies that $Cl^-$ decreases	1
Identifies that equilibrium shifts to left (reactants) to counteract the change	1
Identifies that the blue solution becomes lighter	1

Q25b

2 marks

When a sample of the equilibrium mixture is put into hot water, the mixture turns more blue. Determine whether the forward reaction of the equation is exothermic or endothermic. Explain your reasoning.

Reveal Answer

Adding heat shifts equilibrium towards the endothermic direction and produces $CoCl_4^{2-}$ ions, which are blue.
Therefore, the forward reaction is endothermic.

Marking Criteria

Descriptor	Marks
Identifies that the forward reaction has been favoured	1
Determines that the forward reaction is endothermic	1

Q23

2024

NESA

3 marks

Q23

3 marks

Consider the following equilibrium system.

[\text{Co(H}_2\text{O)}_6]^{2+}(aq) + 4\text{Cl}^-(aq) \rightleftharpoons [\text{CoCl}_4]^{2-}(aq) + 6\text{H}_2\text{O}(l)

$[\text{Co(H}_2\text{O)}_6]^{2+}(aq)$ is pink and $[\text{CoCl}_4]^{2-}(aq)$ is blue. When a solution of these ions and chloride ions is heated, the mixture becomes more blue.

Relate the observed colour change to the change in $K_{eq}$ .

Reveal Answer

The mixture becomes more blue as temperature is increased, which suggests that the concentration of $[\text{CoCl}_4]^{2-}$ is increasing. This means that an increase in temperature favours the forward reaction. This response will result in a larger value of $K_{eq}$ as

K_{eq} = \frac{[[\text{CoCl}_4]^{2-}]}{[[\text{Co}(\text{H}_2\text{O})_6]^{2+}][\text{Cl}^-]^4}

Marking Criteria

Descriptor	Marks
Provides a sound relationship between the observed colour change and $K_{eq}$	3
Provides some information about changes in concentration of reactants and/or products	2
Provides some relevant information	1
None of the above	0

Q11

2025

SCSA

1 mark

Q11

1 mark

Refer to the following equation for the dissolution of gaseous carbon dioxide into the oceans.

$CO_2(g) \leftrightharpoons CO_2(aq) + heat$

Which statement best uses the above equation as part of a justification for why increasing temperatures at the surface of the ocean would lead to a decreased concentration of carbon dioxide in the oceans? Increasing the temperature

increases the rate of the forward reaction and decreases the rate of the reverse reaction leading to an equilibrium shift to the right.

increases the rate of the reverse reaction and decreases the rate of the forward reaction leading to an equilibrium shift to the left.

increases the rate of the endothermic reaction more than the exothermic reaction shifting the equilibrium to the left.

increases the rate of the exothermic reaction more than the endothermic reaction shifting the equilibrium to the left.

Reveal Answer

increases the rate of the forward reaction and decreases the rate of the reverse reaction leading to an equilibrium shift to the right.

Increasing temperature increases the rates of both the forward and reverse reactions due to higher kinetic energy, rather than decreasing one. Additionally, adding heat to an exothermic reaction shifts the equilibrium to the left (reactants), not the right.

increases the rate of the reverse reaction and decreases the rate of the forward reaction leading to an equilibrium shift to the left.

While the equilibrium shifts to the left, increasing temperature increases the rates of both reactions; it does not decrease the rate of the forward reaction.

increases the rate of the endothermic reaction more than the exothermic reaction shifting the equilibrium to the left.

Correct Answer

The reverse reaction is endothermic (absorbs heat). Increasing temperature increases the rate of the endothermic reaction (which has a higher activation energy) more than the exothermic reaction, causing a net shift to the left.

increases the rate of the exothermic reaction more than the endothermic reaction shifting the equilibrium to the left.

Increasing temperature favors the endothermic reaction (the reverse reaction here), not the exothermic reaction. The system shifts to the left to absorb the excess heat.

Q29

2025

NESA

4 marks

Q29

4 marks

Consider the following reaction.

2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) \quad \Delta H = -57.2 \text{ kJ mol}^{-1}

A sealed reaction vessel of fixed volume contains a mixture of $\text{NO}_2$ and $\text{N}_2\text{O}_4$ gases at equilibrium.

Explain the impact of the addition of argon, an inert gas, on the temperature of the system.

Reveal Answer

Addition of argon gas does not change the amount of NO $_2$ or N $_2$ O $_4$ present. As the volume of the system remains consistent, the concentrations of NO $_2$ and N $_2$ O $_4$ are unchanged. Therefore $Q = K_{eq}$ and the system remains at equilibrium. Accordingly, the temperature remains constant.

Marking Criteria

Descriptor	Marks
Provides a thorough explanation of the impact on the temperature of this system	4
Provides an explanation that shows a sound understanding of equilibrium systems	3
Provides a description of an impact on the system	2
Provides some relevant information	1
None of the above	0

2023

QCAA

Paper 1

1 mark

The question refers to the decomposition of hydrogen iodide gas (HI) to produce hydrogen gas ( $\text{H}_2$ ) and iodine gas ( $\text{I}_2$ ) in a sealed 1-litre container.

$2\text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \quad \Delta H = +53.6 \text{ kJ mol}^{-1}$
Colourless Colourless Purple

Identify which change would shift the system from light purple to dark purple.

adding HI(g)

adding a catalyst

decreasing the temperature

increasing the concentration of H $_2$ (g)

Reveal Answer

adding HI(g)

Correct Answer

According to Le Chatelier's Principle, adding the reactant $\text{HI}$ shifts the equilibrium to the right to consume the added substance, resulting in the production of more purple $\text{I}_2$ gas.

adding a catalyst

A catalyst speeds up both the forward and reverse reactions equally but does not change the position of equilibrium or the concentration of $\text{I}_2$ .

decreasing the temperature

The reaction is endothermic ( $\Delta H > 0$ ), so decreasing the temperature shifts the equilibrium to the left (the exothermic direction) to generate heat, consuming $\text{I}_2$ and lightening the color.

increasing the concentration of H $_2$ (g)

Increasing the concentration of the product $\text{H}_2$ shifts the equilibrium to the left to consume the added product, which decreases the concentration of $\text{I}_2$ .

Q17

2023

SCSA

1 mark

Q17

1 mark

In the conversion between sulfur dioxide and sulfur trioxide in a sealed vessel, the following equilibrium is established:

$2\text{ SO}_3\text{(g)} \rightleftharpoons 2\text{ SO}_2\text{(g)} + \text{O}_2\text{(g)} \quad \Delta\text{H} = 198\text{ kJ mol}^{-1}$

For this system, which of the following statements about the equilibrium constant K is correct? K will

increase if the temperature of the system is decreased.

decrease if the partial pressure of $\text{SO}_2\text{(g)}$ is increased.

increase if the temperature of the system is increased.

increase if the pressure of the system is increased.

Reveal Answer

increase if the temperature of the system is decreased.

The reaction is endothermic ( $\Delta\text{H} > 0$ ). Decreasing the temperature shifts the equilibrium to the left, which decreases the equilibrium constant $K$ .

decrease if the partial pressure of $\text{SO}_2\text{(g)}$ is increased.

Changing the partial pressure of a reactant or product shifts the equilibrium position, but it does not change the value of the equilibrium constant $K$ , which is only temperature-dependent.

increase if the temperature of the system is increased.

Correct Answer

Because the reaction is endothermic ( $\Delta\text{H} = 198\text{ kJ mol}^{-1}$ ), increasing the temperature shifts the equilibrium to the right to absorb the added heat, resulting in a higher equilibrium constant $K$ .

increase if the pressure of the system is increased.

Changing the total pressure of the system will shift the equilibrium position (in this case, to the left where there are fewer moles of gas), but it will not change the equilibrium constant $K$ .

2022

SCSA

1 mark

Refer to the following reaction at equilibrium in a closed reaction vessel.

$2\,\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\,\mathrm{SO_3(g)}\ \Delta H = -196\ \mathrm{kJ\ mol^{-1}}$

Which of the following changes will initially decrease the rate at which $\mathrm{SO_2(g)}$ is consumed?

decrease the volume of the reaction vessel

decrease the partial pressure of $\mathrm{O_2(g)}$

heat the reaction vessel

add an appropriate catalyst

Reveal Answer

decrease the volume of the reaction vessel

Decreasing the volume of the vessel increases the concentration of the reactant gases, which increases the rate of the forward reaction.

decrease the partial pressure of $\mathrm{O_2(g)}$

Correct Answer

Decreasing the partial pressure of $\mathrm{O_2(g)}$ lowers its concentration, which directly decreases the rate of the forward reaction where $\mathrm{SO_2(g)}$ is consumed.

heat the reaction vessel

Heating the reaction vessel increases the kinetic energy of the molecules, which increases the rate of both the forward and reverse reactions.

add an appropriate catalyst

Adding a catalyst lowers the activation energy of the reaction, which increases the rate of both the forward and reverse reactions.

Q10

2023

VCAA

1 mark

Q10

1 mark

At constant temperature, which one of the following always corresponds to an increase of the yield of a chemical reaction?

An increase in the value of the equilibrium expression.

An increase in the concentration of the reactants.

A decrease in the activation energy.

An increase in the pressure.

Reveal Answer

An increase in the value of the equilibrium expression.

Correct Answer

The equilibrium expression represents the ratio of products to reactants at equilibrium. An increase in its value means a greater proportion of products is formed, which directly corresponds to a higher yield.

An increase in the concentration of the reactants.

While adding reactants can shift the equilibrium to form more products according to Le Chatelier's principle, it does not change the equilibrium constant and does not always increase the percentage yield.

A decrease in the activation energy.

Decreasing the activation energy (such as by adding a catalyst) increases the rate at which equilibrium is reached, but it does not alter the equilibrium position or the final yield.

An increase in the pressure.

Increasing pressure only increases the yield for reactions that produce fewer moles of gas than they consume. For reactions that produce more moles of gas, an increase in pressure would actually decrease the yield.

2022

QCAA

Paper 1

1 mark

Predict how the system shown will respond when a small amount of aqueous sodium hydroxide is added.

$\text{CH}_3\text{COOH(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{CH}_3\text{COO}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}$

Equilibrium shifts to the left and the pH decreases.

Equilibrium shifts to the right and the pH increases.

Equilibrium shifts to the left and the pH remains the same.

Equilibrium shifts to the right and the pH remains the same.

Reveal Answer

Equilibrium shifts to the left and the pH decreases.

Incorrect. Adding $\text{NaOH}$ introduces hydroxide ions ( $\text{OH}^-$ ) which react with and remove $\text{H}_3\text{O}^+$ . According to Le Chatelier's Principle, removing a product causes the equilibrium to shift to the right, not the left.

Equilibrium shifts to the right and the pH increases.

Incorrect. While adding a base technically causes a slight increase in pH, the system acts as a buffer which resists significant changes in acidity. In the context of ideal buffer behavior, the pH is described as remaining effectively constant.

Equilibrium shifts to the left and the pH remains the same.

Incorrect. The addition of $\text{OH}^-$ consumes $\text{H}_3\text{O}^+$ , creating a stress on the system. To counteract this loss, the equilibrium must shift to the right to produce more $\text{H}_3\text{O}^+$ .

Equilibrium shifts to the right and the pH remains the same.

Correct Answer

Correct. The added $\text{OH}^-$ ions neutralize $\text{H}_3\text{O}^+$ , causing the equilibrium to shift to the right to replace the lost protons. Because the system acts as a buffer, it resists significant changes in pH, keeping it relatively constant.

NESA Chemistry Factors that Affect Equilibrium

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Equilibrium

Economics

Compromise

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NESA Chemistry Factors that Affect Equilibrium

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