Reaction Quotient Q And Gibbs Energy G

ChemistryEquilibriumFor NEET aspirants

The reaction quotient $Q$ has the same algebraic form as the equilibrium constant $K$ but uses concentrations (or partial pressures) at any instant, not just at equilibrium. Comparing $Q$ with $K$ predicts the direction a reaction must shift to reach equilibrium: forward if $Q < K$ , reverse if $Q > K$ , and no net change if $Q = K$ . The thermodynamic link is $Δ G = Δ G^{\circ} + 2.303 R T lo g Q$ , and at equilibrium $Δ G^{\circ} = - 2.303 R T lo g K$ . This connects Le Chatelier reasoning with Gibbs free energy and the Van't Hoff equation.

Key Formulas - Quick Reference

Reaction quotient (general): $Q_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ for $a A + b B ⇌ c C + d D$
Direction rule: $Q < K$ (forward), $Q = K$ (equilibrium), $Q > K$ (reverse)
Non-standard free energy: $Δ G = Δ G^{\circ} + 2.303 R T lo g Q$
Standard free energy and $K$ : $Δ G^{\circ} = - 2.303 R T lo g K = - R T ln K$
Thermodynamic decomposition: $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$
Van't Hoff (integrated form): $lo g \frac{K _{2}}{K _{1}} = \frac{Δ H ^{\circ}}{2.303 R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}]$
Van't Hoff (linear): $lo g K = - \frac{Δ H ^{\circ}}{2.303 R} \cdot \frac{1}{T} + \frac{Δ S ^{\circ}}{2.303 R}$
Spontaneity: $Δ G < 0$ spontaneous, $Δ G = 0$ equilibrium, $Δ G > 0$ non-spontaneous

1. What is the Reaction Quotient Q?

For a general reversible reaction $a A + b B ⇌ c C + d D$ , we can write a ratio with the same form as the equilibrium constant expression but using the concentrations (or partial pressures) present at any instant, not necessarily at equilibrium. This ratio is the reaction quotient $Q$ :

Q_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}

The key contrast with $K_{c}$ is timing. $K_{c}$ uses equilibrium concentrations and is a fixed number at a given temperature. $Q_{c}$ can take any value while the reaction is in progress and evolves as concentrations change, eventually reaching the value $K_{c}$ when equilibrium is established.

2. Qc versus Qp

Just as we have $K_{c}$ (concentrations) and $K_{p}$ (partial pressures for gases), the reaction quotient has two forms:

$Q_{c}$ : uses molar concentrations in $mol L^{- 1}$ . Works for all reactions (solution or gas).
$Q_{p}$ : uses partial pressures in atm. Applies to gas-phase reactions.

The two are related exactly the way $K_{c}$ and $K_{p}$ are: $Q_{p} = Q_{c} (R T)^{Δ n_{g}}$ , where $Δ n_{g}$ counts only gaseous species.

3. Comparing Q with K - Predicting Direction

The comparison of $Q$ with $K$ tells us which way the reaction must shift to reach equilibrium. There are three cases:

Condition	What it means	Which way the reaction shifts
$Q < K$	Too few products relative to equilibrium	Forward (reactants → products)
$Q = K$	System already at equilibrium	No net shift
$Q > K$	Too many products relative to equilibrium	Reverse (products → reactants)

Figure 1: Comparing

Q

with

K

predicts direction -

Q < K

shifts forward,

Q = K

is equilibrium,

Q > K

shifts reverse.

4. Standard Gibbs Free Energy Change ΔG°

The standard Gibbs free energy change $Δ G^{\circ}$ is the free energy change when all reactants and products are in their standard states (1 atm for gases, 1 M for solutions, pure form for solids/liquids). It is a fixed number at a given temperature and reflects the intrinsic thermodynamic driving force.

Its sign predicts whether the reaction, as written, is thermodynamically favourable under standard conditions:

$Δ G^{\circ} < 0$ : forward reaction is spontaneous, $K > 1$ , products favoured.
$Δ G^{\circ} = 0$ : system is at equilibrium under standard conditions, $K = 1$ .
$Δ G^{\circ} > 0$ : forward reaction is non-spontaneous, $K < 1$ , reactants favoured.

5. Non-Standard Free Energy: ΔG = ΔG° + 2.303 RT log Q

Under real (non-standard) conditions, the free energy change depends on the composition of the mixture. The general expression is:

Δ G = Δ G^{\circ} + 2.303 R T lo g Q

This is the master equation of chemical thermodynamics for reactions. It reduces to $Δ G = Δ G^{\circ}$ when $Q = 1$ (standard state), and to $Δ G = 0$ at equilibrium (when $Q = K$ ).

Figure 2: Gibbs free energy

G

against extent of reaction. The system spontaneously moves toward the minimum where

Δ G = 0

and

Q = K

; both forward and reverse approaches reduce

G

6. Relation Between ΔG° and K

At equilibrium $Δ G = 0$ and $Q = K$ . Substituting into the master equation:

0 = Δ G^{\circ} + 2.303 R T lo g K \Rightarrow Δ G^{\circ} = - 2.303 R T lo g K = - R T ln K

This is the single most useful equation connecting thermodynamics to equilibrium. Given $Δ G^{\circ}$ at any temperature you can compute $K$ , and given $K$ from experiment you get $Δ G^{\circ}$ directly.

7. Interpretation of ΔG° Values

$Δ G^{\circ}$	$lo g K$	$K$	Position of equilibrium
Large negative	Large positive	$K ≫ 1$	Very product-favoured
Small negative	Small positive	$K > 1$	Slightly product-favoured
Zero	Zero	$K = 1$	Balanced
Small positive	Small negative	$K < 1$	Slightly reactant-favoured
Large positive	Large negative	$K ≪ 1$	Very reactant-favoured

8. Thermodynamic Decomposition: ΔG° = ΔH° − TΔS°

The standard free energy is built from enthalpy and entropy contributions:

Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}

Combining this with $Δ G^{\circ} = - 2.303 R T lo g K$ gives the temperature dependence of $K$ :

lo g K = - \frac{Δ H ^{\circ}}{2.303 R} \cdot \frac{1}{T} + \frac{Δ S ^{\circ}}{2.303 R}

This is the linear form of the Van't Hoff equation, useful whenever we want to plot $lo g K$ against $1/ T$ .

9. Van't Hoff Equation (Integrated Form)

If we know $K$ at two temperatures and assume $Δ H^{\circ}$ is roughly constant across that range, we can subtract the linear form at $T_{1}$ from the same expression at $T_{2}$ :

lo g \frac{K _{2}}{K _{1}} = \frac{Δ H ^{\circ}}{2.303 R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}]

This is the Van't Hoff equation and is one of the most-used results in equilibrium thermodynamics. It lets us:

Compute $K$ at a new temperature from $K$ at a known temperature.
Extract $Δ H^{\circ}$ from measured equilibrium constants at different temperatures.
Predict qualitatively whether heating shifts equilibrium forward or backward.

10. Van't Hoff Plot: ln K versus 1/T JEE Advanced

Rewriting the linear form using natural logarithms:

ln K = - \frac{Δ H ^{\circ}}{R} \cdot \frac{1}{T} + \frac{Δ S ^{\circ}}{R}

A plot of $ln K$ (y-axis) against $1/ T$ (x-axis) is a straight line with:

Slope $= - \frac{Δ H ^{\circ}}{R}$ → gives standard enthalpy from the slope.
Intercept $= \frac{Δ S ^{\circ}}{R}$ → gives standard entropy from the intercept.

Figure 3: Van't Hoff plot of

ln K

versus

1/ T

. Slope

= - Δ H^{\circ} / R

and intercept

= Δ S^{\circ} / R

. Endothermic reactions have negative slope, exothermic reactions have positive slope.

For an endothermic reaction ( $Δ H^{\circ} > 0$ ), slope is negative, so $ln K$ decreases with $1/ T$ , i.e., $K$ increases as $T$ increases. For an exothermic reaction ( $Δ H^{\circ} < 0$ ), slope is positive, so $K$ decreases as $T$ increases. This is the quantitative form of Le Chatelier's temperature rule.

11. Endothermic vs Exothermic Behaviour

Nature	$Δ H^{\circ}$	Effect of increasing $T$	Van't Hoff slope
Endothermic	$> 0$	$K$ increases, equilibrium shifts forward	Negative
Exothermic	$< 0$	$K$ decreases, equilibrium shifts reverse	Positive

12. Condition for Spontaneity

The sign of the actual $Δ G$ (not $Δ G^{\circ}$ ) determines whether a change is spontaneous under the current conditions:

$Δ G < 0$ : process is spontaneous in the forward direction.
$Δ G = 0$ : system is at equilibrium; no net change occurs.
$Δ G > 0$ : process is non-spontaneous; reverse direction is spontaneous.

Because $Δ G$ depends on composition through $Q$ , a reaction can be non-spontaneous under standard conditions ( $Δ G^{\circ} > 0$ ) but still proceed forward if $Q$ is small enough to make $Δ G < 0$ .

13. Predicting Q From Concentrations

Given a snapshot of concentrations, plug them into the expression for $Q$ exactly as you would for $K$ . If the same reaction has one gaseous participant and one dissolved participant, take the gas in atm and the solute in $mol L^{- 1}$ , matching the way $K$ is defined for the reaction.

14. Applications

The combination of $Q$ , $K$ , and Gibbs free energy underpins several common calculations in JEE and NEET:

Predicting whether a mixture is at equilibrium, and if not, which way it must shift.
Estimating $K$ at a new temperature when $K$ at one temperature and $Δ H^{\circ}$ are known.
Extracting $Δ H^{\circ}$ from an experimental Van't Hoff plot.
Deciding whether a proposed reaction is worth trying: a very positive $Δ G^{\circ}$ means $K$ is tiny and yield will be poor even with clever conditions.

Solved Example 1

For the equilibrium

A (g) + B (g) ⇌ 2 C (g)

at 25 °C, a 2 L vessel contains 1 mol

A

, 2 mol

B

, and 3 mol

C

. Predict the direction of shift if (a)

K_{c} = 3

, (b)

K_{c} = 6

, (c)

K_{c} = 4.5

Solution:

Concentrations: $[A] = 0.5 M$ , $[B] = 1.0 M$ , $[C] = 1.5 M$ .

$Q_{c} = \frac{[ C ] ^{2}}{[ A ] [ B ]} = \frac{( 1.5 ) ^{2}}{( 0.5 ) ( 1.0 )} = \frac{2.25}{0.5} = 4.5$

(a) $Q_{c} = 4.5 > K_{c} = 3$ → reaction shifts reverse.
(b) $Q_{c} = 4.5 < K_{c} = 6$ → reaction shifts forward.
(c) $Q_{c} = K_{c} = 4.5$ → system is at equilibrium, no net shift.

Solved Example 2

For

N O B r (g) ⇌ N O (g) + \frac{1}{2} B r_{2} (g)

K_{p} = 0.15

atm at 90 °C. A mixture at 90 °C has

p_{N O B r} = 0.5

atm,

p_{N O} = 0.4

atm,

p_{B r_{2}} = 0.2

atm. Is

B r_{2}

produced or consumed?

Solution:

$Q_{p} = \frac{p _{N O} \cdot ( p _{B r_{2}} ) ^{1/2}}{p _{N O B r}} = \frac{0.4 \times ( 0.2 ) ^{1/2}}{0.5} = \frac{0.4 \times 0.447}{0.5} \approx 0.358$

$Q_{p} = 0.358 > K_{p} = 0.15$ , so the reaction shifts reverse. $B r_{2}$ is consumed.

Solved Example 3

The rate of consumption of

B

A (a q) + B (a q) ⇌ C (a q) + D (a q)

at 25 °C obeys

- d [B] / d t = 4 \times 1 0^{- 4} [A] [B] - 1.33 \times 1 0^{- 5} [C] [D]

. In a mixture with

[A] = [B] = [C] = [D] = 1

M, is the reaction spontaneous forward?

Solution:

$K = k_{f} / k_{b} = 4 \times 1 0^{- 4} /1.33 \times 1 0^{- 5} = 30$ .
$Q = [C] [D] / [A] [B] = 1$ .
Since $Q = 1 < K = 30$ , the reaction proceeds spontaneously in the forward direction.

Solved Example 4

For

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

K_{p} = 1.6 \times 1 0^{- 4}

atm^{- 2}

at 400 °C. Given

Δ H^{\circ} = - 25.14

kcal mol

^{- 1}

, find

K_{p}

at 500 °C.

Solution:

$T_{1} = 673$ K, $T_{2} = 773$ K, $R = 2$ cal mol $^{- 1}$ K $^{- 1}$ .

$lo g \frac{K _{2}}{K _{1}} = \frac{Δ H ^{\circ}}{2.303 R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}] = \frac{- 25140}{2.303 \times 2} [\frac{773 - 673}{773 \times 673}]$

$= - 5459 \times \frac{100}{520229} \approx - 1.05$

$lo g K_{2} = lo g (1.6 \times 1 0^{- 4}) - 1.05 = - 3.80 - 1.05 = - 4.85$

$K_{2} \approx 1.4 \times 1 0^{- 5}$ $atm^{- 2}$ . As expected for an exothermic reaction, $K$ decreases as $T$ rises.

Solved Example 5

The equilibrium constant of a reaction at 298 K is

K = 4.0 \times 1 0^{6}

. Compute

Δ G^{\circ}

at this temperature.

Solution:

$Δ G^{\circ} = - 2.303 R T lo g K = - 2.303 \times 8.314 \times 298 \times lo g (4 \times 1 0^{6})$

$lo g (4 \times 1 0^{6}) = 6.602$ ; $2.303 \times 8.314 \times 298 \approx 5706$ J mol $^{- 1}$ .

$Δ G^{\circ} = - 5706 \times 6.602 \approx - 37671$ J mol $^{- 1} = - 37.7$ kJ mol $^{- 1}$ .

Strongly negative $Δ G^{\circ}$ confirms the reaction is strongly product-favoured.

Solved Example 6

For

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

at 1000 K,

K_{c} = 2.24 \times 1 0^{22}

. Is the forward reaction favoured, and estimate

Δ G^{\circ}

Solution:

$K ≫ 1$ , so the forward reaction is strongly product-favoured.

$Δ G^{\circ} = - 2.303 \times 8.314 \times 1000 \times lo g (2.24 \times 1 0^{22})$

$lo g (2.24 \times 1 0^{22}) = 22.35$ ; $2.303 \times 8.314 \times 1000 = 19145$ J mol $^{- 1}$ .

$Δ G^{\circ} = - 19145 \times 22.35 \approx - 4.28 \times 1 0^{5}$ J mol $^{- 1} = - 428$ kJ mol $^{- 1}$ .

Solved Example 7

Variation of

lo g K

with temperature follows

lo g K = lo g A - \frac{Δ H ^{\circ}}{2.303 R T}

. A plot of

lo g K

against

1/ T

is a straight line with slope

- tan θ

where

tan θ = 0.5

, and vertical intercept 10. Find (a)

Δ H^{\circ}

, (b) pre-exponential factor

A

, (c)

K

at 300 K.

Solution:

(a) Slope $= - \frac{Δ H ^{\circ}}{2.303 R} = - 0.5$ , so $Δ H^{\circ} = 0.5 \times 2.303 \times 8.314 \approx 9.57$ J mol $^{- 1}$ .

(b) Intercept $= lo g A = 10$ , so $A = 1 0^{10}$ .

(c) $lo g K = 10 - \frac{9.57}{2.303 \times 8.314 \times 300} = 10 - 0.00166 \approx 9.998$ , so $K \approx 9.96 \times 1 0^{9}$ .

Solved Example 8

For the reaction

H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)

K_{c} = 50

at 700 K. In a 1 L flask, 1 mol

H_{2}

, 1 mol

I_{2}

, and 1 mol

H I

are mixed at 700 K. Which way does the reaction proceed?

Solution:

$Q_{c} = \frac{[ H I ] ^{2}}{[ H _{2} ] [ I _{2} ]} = \frac{1 ^{2}}{1 \times 1} = 1$ .

Since $Q_{c} = 1 < K_{c} = 50$ , the reaction shifts forward to produce more $H I$ until $Q_{c}$ rises to 50.

Common Mistakes to Avoid

Watch out

Confusing $Q$ with $K$ : $Q$ is calculated from current concentrations at any moment; $K$ is only defined at equilibrium.
Using natural log when the formula asks for common log, or vice versa. Check whether the factor 2.303 is present.
Substituting $Δ H^{\circ}$ in kJ but $R$ in J mol $^{- 1}$ K $^{- 1}$ . Both must be in the same energy unit.
Forgetting to include the sign of $Δ H^{\circ}$ in the Van't Hoff equation. Exothermic reactions have $Δ H^{\circ} < 0$ .
Reading the slope of the Van't Hoff plot as $Δ H^{\circ} / R$ instead of $- Δ H^{\circ} / R$ . The minus sign matters.
Confusing $Δ G$ with $Δ G^{\circ}$ . $Δ G^{\circ}$ is fixed at a given temperature; $Δ G$ depends on composition via $Q$ .
Treating $Δ G^{\circ} > 0$ as "reaction cannot happen". It just means $K < 1$ ; the forward reaction is still possible if $Q$ is small enough.
Including pure solids or pure liquids in the expression for $Q$ or $K$ . Their activities are taken as 1.

Frequently Asked Questions

Q1. What is the difference between the reaction quotient Q and the equilibrium constant K?

Both have the same algebraic form as a ratio of product concentrations (or partial pressures) to reactant concentrations, each raised to the appropriate stoichiometric power. The difference is timing: $K$ uses concentrations only when the system is at equilibrium and is a fixed number at a given temperature, while $Q$ uses concentrations at any moment during the reaction and changes over time until it equals $K$ .

Q2. How do I use Q to predict the direction of a reaction?

Calculate $Q$ using the current concentrations and compare it with $K$ at the same temperature. If $Q < K$ , the reaction proceeds forward to make more products. If $Q > K$ , it proceeds in reverse to consume products and regenerate reactants. If $Q = K$ , the system is already at equilibrium and no net change occurs.

Q3. What does ΔG° = 0 mean?

When the standard Gibbs free energy change is zero, the equilibrium constant $K$ equals 1, meaning products and reactants are equally favoured under standard conditions. It does not mean the reaction is stuck; it just means neither direction has a thermodynamic advantage when all species are at their standard states.

Q4. Why does K depend on temperature but not on concentration or pressure?

The equilibrium constant is fixed by $Δ G^{\circ}$ , which changes only with temperature since $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ . Changes in concentration, pressure, volume, or catalyst shift the position of equilibrium (they change $Q$ ), but the value of $K$ itself only responds to temperature changes.

Q5. How is the Van't Hoff equation derived?

Start from $Δ G^{\circ} = - 2.303 R T lo g K$ and $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ . Equate them and divide by $- 2.303 R T$ to get $lo g K = - Δ H^{\circ} / (2.303 R T) + Δ S^{\circ} / (2.303 R)$ . Writing this at two temperatures $T_{1}$ and $T_{2}$ and subtracting eliminates the entropy term, giving the two-point form.

Q6. What does the Van't Hoff plot slope tell us?

The slope of $ln K$ against $1/ T$ is $- Δ H^{\circ} / R$ . A negative slope means the reaction is endothermic; a positive slope means it is exothermic. The intercept $Δ S^{\circ} / R$ gives the entropy change. This makes Van't Hoff plots a powerful experimental route to $Δ H^{\circ}$ and $Δ S^{\circ}$ .

Q7. Is a reaction with a positive ΔG° impossible?

No. A positive $Δ G^{\circ}$ means the equilibrium constant is less than 1, so at standard conditions the reactants are favoured. But the actual $Δ G$ depends on composition through $Δ G = Δ G^{\circ} + 2.303 R T lo g Q$ . If $Q$ is small enough (few products present), $Δ G$ can still be negative and the forward reaction proceeds.

Q8. Does temperature affect ΔG° the same way it affects K?

Yes, but through different windows. $Δ G^{\circ}$ varies with $T$ mainly through the $- T Δ S^{\circ}$ term, while $K$ varies with $T$ through the Van't Hoff relation, which is essentially the same equation rearranged. Both variations reflect the shift in thermodynamic favourability as temperature changes.

Q9. Why does a catalyst not change K or ΔG°?

A catalyst lowers the activation energy of both the forward and reverse reactions by the same amount, so the rates rise together and the system reaches equilibrium faster. It does not shift the position of equilibrium or the value of $K$ , because $K$ is set by the thermodynamics of the reactants and products, not by the path connecting them.

Previous year questions on Reaction Quotient Q And Gibbs Energy G

3 questions from past papers, each with a step-by-step solution.

Ready to master Equilibrium?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test