What is the equilibrium constant?

For a general reaction aA + bB cC + dD, the equilibrium constant is

TASChemistrySyllabus dot point

How do reversible reactions reach and respond to disturbances of equilibrium?

Explain dynamic equilibrium, write equilibrium constant expressions, and predict shifts using Le Chatelier's principle.

Dynamic equilibrium, the equilibrium constant Kc, the reaction quotient Q, and using Le Chatelier's principle to predict how concentration, pressure and temperature changes shift a system.

Generated by Claude Opus 4.77 min answerUpdated 2026-05-29

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What this dot point is asking

You need to describe what equilibrium is at the particle level, write and use the equilibrium constant expression, and predict the direction a system shifts when it is disturbed.

Dynamic equilibrium

A reversible reaction can proceed forwards and backwards. Early on the forward rate is high because reactant concentration is high. As products build up, the reverse rate rises. Eventually the two rates become equal and the concentrations of all species stop changing. This is dynamic equilibrium: both reactions are still occurring, but there is no net change. It can only be reached in a closed system, and it can be approached from either direction.

The equilibrium constant

For a general reaction $aA + bB \rightleftharpoons cC + dD$ , the equilibrium constant is

Square brackets are equilibrium concentrations in $\text{mol L}^{-1}$ . Pure solids and pure liquids are left out because their concentration is effectively constant. A large $K_c$ (much greater than 1) means products are favoured; a small $K_c$ means reactants are favoured. $K_c$ depends only on temperature, so changing concentration or pressure does not change its value.

The reaction quotient Q

$Q$ has the same form as $K_c$ but uses concentrations at any instant, not just at equilibrium. Comparing $Q$ with $K_c$ tells you which way a reaction must proceed:

If $Q < K_c$ , the forward reaction is favoured (more product forms).
If $Q > K_c$ , the reverse reaction is favoured.
If $Q = K_c$ , the system is at equilibrium.

Le Chatelier's principle

If a system at equilibrium is disturbed, it shifts to partly counteract the disturbance.

Increasing the concentration of a reactant shifts the position to the right; removing product also shifts it right.
For gaseous systems, increasing pressure (decreasing volume) shifts the system towards the side with fewer moles of gas.
Increasing temperature favours the endothermic direction. For an exothermic forward reaction ( $\Delta H < 0$ ), heating shifts the system left and decreases $K_c$ .
A catalyst speeds up both forward and reverse rates equally, so it shortens the time to reach equilibrium but does not change the position or $K_c$ .

When you answer a Le Chatelier question, always state the change, the direction of the shift, and the effect on the named species or on $K_c$ . Markers reward that three-part structure.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 TASC10 marksNitrogen monoxide, oxygen and nitrogen dioxide reach equilibrium: 2NO(g) + O2(g) reversible 2NO2(g), delta H = -114 kJ. For an increase in temperature and for removal of NO2, state the effect on the initial reaction quotient (Q), the equilibrium constant (Kc) and the amount of NO. Then state what conditions of temperature and pressure give the highest yield of NO2 and explain.

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Increase in temperature: Q is unchanged at the instant of the change (concentrations have not yet altered); Kc decreases (the forward reaction is exothermic, so raising temperature favours the reverse reaction); the amount of NO increases as the system shifts back toward reactants.

Removal of NO2: Q decreases (removing product lowers the numerator); Kc is unchanged (Kc depends only on temperature); the amount of NO decreases as the forward reaction is favoured to replace the NO2. (6 marks for the completed table.)

Highest yield of NO2 (4 marks): use low temperature and high pressure. Low temperature favours the exothermic forward reaction (more NO2). High pressure favours the side with fewer gas moles - the forward reaction goes from 3 mol of gas to 2 mol - so it also increases NO2. (A practical compromise temperature is needed for an acceptable rate, but for maximum yield alone, low temperature and high pressure.)

2022 TASC4 marksConsider the equilibrium 2NOCl(g) reversible 2NO(g) + Cl2(g). Predict which reaction (forward or reverse) favours maximum entropy and explain. Then predict whether the forward reaction is exothermic or endothermic and explain.

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Maximum entropy (2 marks). The forward reaction increases the number of gas molecules from 2 mol to 3 mol, increasing the disorder (positional entropy) of the system. So the forward reaction favours maximum entropy.

Exothermic or endothermic (2 marks). Breaking the N-Cl bonds in NOCl requires energy and the products are separate NO and Cl2 molecules; the forward reaction is bond-breaking dominated and is endothermic. (Equivalently, this dissociation only proceeds significantly at high temperature, consistent with an endothermic forward reaction whose equilibrium shifts right as temperature rises.)

2021 TASC2 marksMethanol is produced by CO(g) + 2H2(g) reversible CH3OH(g), delta H = -91 kJ mol-1. Use Le Chatelier's principle to explain the advantage of using a higher pressure on the equilibrium yield of methanol.

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Increasing the pressure puts stress on the system, which responds by shifting in the direction that reduces the total number of gas molecules.

The forward (left) side has 3 mol of gas (1 CO + 2 H2) while the right side has only 1 mol of gas (CH3OH). To partially relieve the increased pressure the equilibrium shifts toward the side with fewer gas molecules, that is, toward the products.

So a higher pressure increases the equilibrium yield of methanol. (2 marks: relate pressure to mole difference and the resulting forward shift.)

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