Equilibrium · Lesson 5 of 14

Using K, Q, and Gibbs Energy

“Predict direction, calculate equilibrium composition, and connect the result to Gibbs energy.”

Learning Objectives

• Use the magnitude of K as an indication of reaction extent without confusing it with speed. • Predict a net reaction direction by comparing Q with K. • Construct initial–change–equilibrium tables and reject impossible mathematical roots. • Relate actual and standard Gibbs-energy changes to Q and K.

Extent and speed answer different questions

A large equilibrium constant indicates that the product-to-reactant activity ratio is large at equilibrium for the written reaction. A small constant indicates the opposite. The chapter uses values above about a thousand and below about one-thousandth as useful broad categories. These are guides, not sharp physical boundaries: stoichiometry, concentrations and the way an equation is written still matter.

Hydrogen reacting with oxygen or chlorine can have a very large equilibrium constant, while formation of nitric oxide from nitrogen and oxygen at room temperature has a very small one. A constant in between often allows substantial quantities of both sides. None of these statements tells us how quickly equilibrium is reached. A reaction with a large K can remain extremely slow without a suitable pathway.

Extent, direction and rate are different questions— K describes the equilibrium ratio; Q describes the current mixture. Reaction speed requires kinetic information.

Use the reaction quotient for the mixture you have now

The reaction quotient Q uses exactly the same product-over-reactant expression as K but is evaluated with the current activities or their suitable approximations. Those values need not be equilibrium values. A mixture with Q below K needs a larger product-to-reactant ratio, so its net change is forward. A mixture with Q above K moves in reverse.

Direction testLaTeX

Q < K : net forward; Q = K : equilibrium; Q > K : net reverse

Use the same written equation, temperature and standard-state convention for Q and K.

For $H_{2} + I_{2} ⇌ 2HI$ , concentrations 0.10 M, 0.20 M and 0.40 M give $Q_{c} = 0.4 0^{2} / (0.10 \times 0.20) = 8.0$ . If Kc is 57.0 at that temperature, more HI forms. Q increases towards K as the composition changes; K itself does not change during this fixed-temperature relaxation.

Example — A quotient larger than the constant

Problem
For $2A ⇌ B + C$ , $K_{c} = 2.0 \times 1 0^{- 3}$ . At one instant all three concentrations are $3.0 \times 1 0^{- 4} M$ . Predict the net direction.

1.Construct $Q_{c} = [B] [C] / [A]^{2}$ .
2.All three concentrations have the same numerical value, so $Q_{c} = 1$ .
3.Because $1 > 0.002$ , B and C are consumed overall to form A. Equal species concentrations do not imply equilibrium.

Use stoichiometry before solving the equilibrium equation

When only initial values and K are known, define a reaction change x and express every concentration in terms of it. This is often called an initial–change–equilibrium table. The change row follows coefficients in the balanced equation, so it protects atom conservation. Solve only after this table is correct.

A reliable sequence is to balance the equation, list initial values, predict the direction if necessary, write stoichiometric changes, substitute final expressions in K, solve, and check. Concentrations must be nonnegative. A root larger than the available amount of reactant is mathematically possible but chemically impossible. Substituting the accepted root back into K is a valuable independent check.

Example — Pressure data for nitrogen dioxide formation

Problem
A 1.00 L vessel contains 13.8 g of N2O4 at 400 K. It reaches $N_{2} O_{4} ⇌ 2N O_{2}$ equilibrium at total pressure 9.15 bar. Using molar mass 92.0 g per mole and R = 0.083 bar litre per mole kelvin, find the partial pressures, Kp and Kc.

1.The starting amount is $13.8/92.0 = 0.150 mol$ . Its initial pressure is $0.150 (0.083) (400) /1.00 = 4.98 bar$ .
2.If N2O4 loses x bar of partial pressure, NO2 gains 2x bar. Total pressure is $4.98 - x + 2 x = 9.15$ , so $x = 4.17 bar$ .
3.Thus $p_{N_{2} O_{4}} = 0.81 bar$ and $p_{N O_{2}} = 8.34 bar$ . Their sum checks the total.
4.The numerical constant is $K_{p} = 8.3 4^{2} /0.81 = 85.87$ . Since the gas-mole change is +1, $K_{c} = 85.87/ (0.083 \times 400) = 2.59$ . These results follow the supplied dataset and rounding convention.

Example — Selecting a chemical root

Problem
Initially 3.00 mol of PCl5 is placed in a 1.00 L vessel. For $PC l_{5} ⇌ PC l_{3} + C l_{2}$ at 380 K, Kc = 1.80. Find the equilibrium composition.

1.Initial concentrations are 3.00 M, 0 and 0. If x M dissociates, equilibrium concentrations are $3.00 - x$ , x and x.
2.Substitution gives $x^{2} / (3.00 - x) = 1.80$ , so $x^{2} + 1.80 x - 5.40 = 0$ .
3.The roots are $(- 1.80 \pm 24.84) /2$ . The positive root is 1.592 M; the negative root is rejected.
4.Therefore $[PC l_{5}] = 1.408 M$ and $[PC l_{3}] = [C l_{2}] = 1.592 M$ . Check: $1.59 2^{2} /1.408 \approx 1.80$ .

Gibbs energy explains the quotient test

At fixed temperature and pressure, the actual Gibbs-energy change for the written reaction tells us the spontaneous net direction. A negative reaction Gibbs energy favours forward change, a positive value favours reverse change, and zero identifies equilibrium. The standard reaction Gibbs energy refers to standard-state activities; it is not generally the actual value in the current mixture.

Actual reaction Gibbs energyLaTeX

Δ_{r} G = Δ_{r} G^{\circ} + R T ln Q

R = 8.314 joules per mole kelvin when Gibbs energies are in joules per mole. T is absolute temperature and Q is dimensionless.

At equilibrium, the actual reaction Gibbs energy is zero and Q equals K. Substitute those two facts and rearrange: the standard reaction Gibbs energy is the negative of RT times the natural logarithm of K. Combining the two equations also gives a direct form in terms of Q/K.

Equilibrium and standard Gibbs energyLaTeX

Δ_{r} G^{\circ} = - R T ln K, K = e^{- Δ_{r} G^{\circ} / (R T)}, Δ_{r} G = R T ln (\frac{Q}{K})

A negative standard value means K > 1, a positive value means K < 1, and zero means K = 1. The actual direction still depends on Q.

Example — A small equilibrium constant

Problem
For a supplied model of uncoupled glucose phosphorylation, $Δ_{r} G^{\circ} = + 13.8 kJ mo l^{- 1}$ at 298 K. Find K.

1.Convert the energy to $13800 J mo l^{- 1}$ so it matches the units of R.
2.Calculate $ln K = - 13 800/ (8.314 \times 298) = - 5.570$ .
3.Exponentiation gives $K \approx 3.81 \times 1 0^{- 3}$ . This small standard equilibrium ratio does not by itself specify the direction in an arbitrary starting mixture.

Example — Standard Gibbs energy for sucrose hydrolysis

Problem
For $sucrose + water ⇌ glucose + fructose$ , a supplied equilibrium constant is $2.0 \times 1 0^{13}$ at 300 K. Calculate the standard reaction Gibbs energy.

1.Use $Δ_{r} G^{\circ} = - R T ln K$ . Water is treated consistently with the constant supplied for this aqueous model.
2.Evaluate $ln (2.0 \times 1 0^{13}) = ln 2 + 13 ln 10 \approx 30.627$ .
3.Then $Δ_{r} G^{\circ} = - (8.314) (300) (30.627) = - 7.64 \times 1 0^{4} J mo l^{- 1} = - 76.4 kJ mo l^{- 1}$ .

Common mistake

K greater than one does not guarantee forward change for every starting mixture. A product-rich mixture can have Q greater than K and react in reverse. Also, spontaneous means thermodynamically favoured under the conditions; it does not mean rapid.

Quick check

If Q is below K, the net change for the written reaction is

Quick check

A large K tells us that

Quick check

Which root must be rejected in an equilibrium concentration calculation?

Quick check

At equilibrium, which statement is correct?

Quick check

If standard reaction Gibbs energy is positive, K is

Quick check

A mixture has Q = 100 and K = 10. Its actual reaction Gibbs energy is

Practice Problems

For A + B forming C, concentrations are 0.20, 0.50 and 0.30 M and Kc = 5.0. Predict the net direction.
For A forming B with Kc = 4.0, start with 1.0 M A and no B. Find both equilibrium concentrations.
Explain why a negative quadratic root may be chemically impossible even though the algebra is correct.
Calculate standard reaction Gibbs energy at 298 K for K = 10.0.
A reaction has K = 0.010 but begins with Q = 0.0001. Predict its direction and the sign of its actual reaction Gibbs energy.
Explain how a slow reaction can have a very large equilibrium constant.

Key Takeaways

K indicates the equilibrium ratio; Q tells us whether the current mixture must change. Stoichiometric change tables conserve atoms and reveal the physically allowable solutions. The Gibbs-energy relations explain the direction test and connect equilibrium with thermodynamics.

Previous · Lesson 4

Gas Pressures and Heterogeneous Equilibria

Next · Lesson 6

Changing an Equilibrium

Extent and speed answer different questions

Use the reaction quotient for the mixture you have now

Use stoichiometry before solving the equilibrium equation

Gibbs energy explains the quotient test

Quiz

Practice Problems

Key Takeaways