Midterm 1 : Equilibrium

Distinguish homogeneous and heterogeneous equilibria and write equilibrium constants for both types of reactions from a balanced equation.

Definitions

• A homogeneous equilibrium has all reactants and products in the same phase, such as all gases or all dissolved in one solution.

  • Example: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$.

• A heterogeneous equilibrium involves more than one phase, such as a solid in contact with a gas or a solution.

  • Example: $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$.

Why the distinction matters: activities

• The rigorous equilibrium constant is built from activities, $\mathbf{a_i}$, which are dimensionless:

  • $\mathbf{K = \prod_i a_i^{\nu_i}}$, with $\nu_i > 0$ for products and $\nu_i < 0$ for reactants.

• Activity of an ideal gas: $a = p/p^\circ$.

• Activity of a dilute solute: $a = c/c^\circ$.

• Activity of a pure solid or pure liquid: $a = 1$.

  • Its molar free energy is essentially fixed, because the molecular environment does not change as more or less of it is present.

• Activity of a solvent in a dilute solution: $a \approx 1$, so water is omitted.

• Because $K$ is a product of dimensionless activities, $K$ is dimensionless. In practice we write concentrations or pressures and implicitly divide by $c^\circ$ or $p^\circ$.

Writing the expression

• Products go in the numerator and reactants in the denominator, with each activity raised to its stoichiometric coefficient.

• Omit pure solids and liquids.

• Pressure form: $K_p$ (gases). Concentration form: $K_c$. Mixed systems use whichever variable applies to each species.

Reaction	Expression
$\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ (all gas)	$$K = \dfrac{p_{\text{NH}_3}^2}{p_{\text{N}_2}\,p_{\text{H}_2}^3}$$
$$\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$$	$$K = p_{\text{CO}_2}$$
$$\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq)$$	$$K = [\text{Pb}^{2+}][\text{Cl}^-]^2$$
$$\text{CH}_3\text{COOH} + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+ + \text{CH}_3\text{COO}^-$$	$$K = \dfrac{[\text{H}_3\text{O}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$$
$\tfrac12\text{H}_2(g) + \tfrac12\text{I}_2(s) \rightleftharpoons \text{HI}(g)$ (your slide)	$$K = \dfrac{p_{\text{HI}}}{p_{\text{H}_2}^{1/2}}$$

Common pitfalls

• A solid does not appear in $K$, but it must be present for equilibrium to exist. If all the $\text{CaCO}_3$ decomposes, the system is not at equilibrium and $p_{\text{CO}_2}$ is no longer fixed.

• For $\text{CaCO}_3(s)$, $p_{\text{CO}_2}$ depends only on temperature, not on how much solid is present.

• Do not leave out coefficients. Writing $[\text{Pb}^{2+}][\text{Cl}^-]$ instead of $[\text{Pb}^{2+}][\text{Cl}^-]^2$ is the most common error.

• Water is omitted only when it is the solvent in dilute solution. If water is a gaseous reactant or product, such as $\text{H}_2\text{O}(g)$, it must be included.

Worked example
Write $K$ for $2\text{NaHCO}_3(s) \rightleftharpoons \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)$.

1. The phases are two solids and two gases, so the system is heterogeneous.

2. Both solids have $a = 1$ and drop out.

3. The gases appear in the numerator with coefficient 1:

   • $K = p_{\text{H}_2\text{O}} \cdot p_{\text{CO}_2}$, which is dimensionless after dividing each pressure by $p^\circ$.

Relate the Gibbs free energy of reaction to the composition of the reaction mixture.

Core relationship

• $\Delta G = \Delta G^\circ + RT\ln Q$

  • $\Delta G$ (J mol⁻¹) is the Gibbs free energy change per mole of reaction at the current composition.

  • $\Delta G^\circ$ (J mol⁻¹) is the standard value, with every species in its standard state.

  • $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$.

  • $T$ is in kelvin.

  • $Q$ is the reaction quotient, with the same form as $K$ but using current activities.

Where it comes from

• Each species has a chemical potential $\mu_i = \mu_i^\circ + RT\ln a_i$.

  • The $RT\ln a_i$ term reflects the entropy of mixing: a more dilute species has more available microstates, so it has lower free energy per mole.

• The reaction free energy is $\Delta G = \sum_i \nu_i \mu_i$.

  • Substituting gives $\Delta G = \sum \nu_i\mu_i^\circ + RT\sum \nu_i \ln a_i = \Delta G^\circ + RT\ln Q$.

• $\Delta G$ is the slope of the Gibbs energy versus extent of reaction at constant $T$ and $p$.

  • If $\Delta G < 0$, $G$ falls as the reaction moves forward, so the forward reaction is spontaneous.

  • If $\Delta G > 0$, the reverse reaction is spontaneous.

  • At the minimum of $G$, the slope is zero: $\Delta G = 0$, which is equilibrium.

Connection to the second law (from your slides)

• At constant $T$ and $p$ with a large heat bath, $\Delta G_{sys} = -T\Delta S_{univ}$.

  • $\Delta S_{surr} = -\Delta H_{sys}/T$, so $-\Delta G_{sys}/T = \Delta S_{sys} + \Delta S_{surr}$.

• A negative $\Delta G$ is therefore the same statement as $\Delta S_{univ} > 0$.

Interpreting the sign of $\Delta G$

• $Q < K$ gives $\ln Q < \ln K$ and $\Delta G < 0$, so the reaction goes forward.

• $Q = K$ gives $\Delta G = 0$, so the system is at equilibrium.

• $Q > K$ gives $\Delta G > 0$, so the reaction goes in reverse.

Pitfalls

• $\Delta G^\circ$ and $\Delta G$ are different things. $\Delta G^\circ$ is a fixed number at a given $T$. $\Delta G$ changes as the reaction proceeds and reaches zero at equilibrium.

• $\Delta G^\circ > 0$ does not mean no reaction occurs. If $Q$ is small enough, $\Delta G$ can still be negative.

• $\Delta G$ gives the direction and driving force, not the rate. Kinetics decides how fast.

• Use $\Delta G$ in J if $R$ is in J. Mixing kJ and J is the classic error.

• $Q$ uses $p/p^\circ$ and $c/c^\circ$, and pure solids and liquids are omitted, as for $K$.

Worked example (your slide’s NH₃ case)
$\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ at 298 K, with $p_{\text{N}_2} = 1.0$, $p_{\text{H}_2} = 3.0$, $p_{\text{NH}_3} = 4.0$ atm and $\Delta G^\circ = -33.32\ \text{kJ mol}^{-1}$.

1. $Q = \dfrac{4.0^2}{(3.0)^3(1.0)} = \dfrac{16}{27} = 0.593$.

2. $RT = (8.314\ \text{J mol}^{-1}\text{K}^{-1})(298\ \text{K}) = 2478\ \text{J mol}^{-1} = 2.478\ \text{kJ mol}^{-1}$.

3. $RT\ln Q = (2.478)(\ln 0.593) = (2.478)(-0.523) = -1.30\ \text{kJ mol}^{-1}$.

4. $\Delta G = -33.32 + (-1.30) = \mathbf{-34.6\ kJ\ mol^{-1}}$, or about $-35$ kJ mol⁻¹ to 2 sig. figs.

   • Your slide shows $-1.43$ for the correction term, but the arithmetic gives $-1.30$. The final answer to 2 sig. figs. is unchanged.

5. $\Delta G < 0$, so the mixture shifts forward toward NH₃.

Calculate an equilibrium constant from a standard Gibbs free energy.

Derivation

• Start with $\Delta G = \Delta G^\circ + RT\ln Q$.

• At equilibrium, $\Delta G = 0$ and $Q = K$:

  • $0 = \Delta G^\circ + RT\ln K$.

• Therefore $\Delta G^\circ = -RT\ln K$, or $\ln K = -\dfrac{\Delta G^\circ}{RT}$, or $K = e^{-\Delta G^\circ/RT}$.

Variables and units

• $\Delta G^\circ$ is in J mol⁻¹ (convert kJ to J).

• $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$.

• $T$ is in K.

• $K$ is dimensionless.

• $\Delta G^\circ$ is per mole of reaction as written, so it is tied to the specific balanced equation.

Obtaining $\Delta G^\circ$

• From formation values: $\Delta G^\circ_{rxn} = \sum n\,\Delta G^\circ_f(\text{products}) - \sum m\,\Delta G^\circ_f(\text{reactants})$.

  • Elements in their standard states have $\Delta G_f^\circ = 0$.

• From enthalpy and entropy: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$.

• From Hess’s law: combine reactions and add their $\Delta G^\circ$ values.

Physical interpretation

• $\Delta G^\circ$ measures the free-energy gap between pure standard-state reactants and pure standard-state products.

  • A large negative value makes products much more stable, so $K \gg 1$.

  • A large positive value gives $K \ll 1$.

• $\Delta G^\circ = 0$ gives $K = 1$.

• At 298 K, every change of $2.303RT = 5.71\ \text{kJ mol}^{-1}$ in $\Delta G^\circ$ changes $K$ by a factor of 10.

Pitfalls

• $K$ depends exponentially on $\Delta G^\circ$, so small errors in $\Delta G^\circ$ give large errors in $K$.

• Your slide says “the higher the $\Delta G^\circ_{rxn}$, the more likely the reaction.” This is backwards. More negative $\Delta G^\circ$ means larger $K$ and a more product-favored reaction.

• $K > 1$ does not mean a fast reaction.

• $\Delta G^\circ$ is evaluated at the specific temperature of interest.

• The sign of $\Delta G^\circ$ tells you which side is favored at standard conditions, not whether the reaction proceeds under non-standard conditions. That is the job of $\Delta G$.

Worked example (your slide)
$\tfrac12\text{H}_2(g) + \tfrac12\text{I}_2(s) \rightarrow \text{HI}(g)$, $\Delta G^\circ = +1.70\ \text{kJ mol}^{-1}$ at 25.00 °C.

1. Convert: $\Delta G^\circ = 1700\ \text{J mol}^{-1}$ and $T = 298.15\ \text{K}$.

2. $\ln K = -\dfrac{1700\ \text{J mol}^{-1}}{(8.3145\ \text{J mol}^{-1}\text{K}^{-1})(298.15\ \text{K})} = -0.686$. The mol and K units cancel, leaving a dimensionless number.

3. $K = e^{-0.686} = \mathbf{0.50}$.

4. Because $\Delta G^\circ$ is positive, $K < 1$, which is consistent.

For scale, the slide’s $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}(g)$ has $\Delta G^\circ = -457.14$ kJ. Then $\ln K = 457140/2478 = 184.5$ and $K = 1.4\times10^{80}$.

Calculate an equilibrium concentration.

Strategy

• Step 1: Write the balanced equation and the correct $K$ expression, omitting pure solids and liquids (Card 1).

• Step 2: Make sure $K$ matches the equation as written and the temperature of interest (Card 7).

• Step 3: Decide which quantities are known at equilibrium and which are unknown.

• Step 4: Solve algebraically for the unknown.

  • If only one concentration is unknown, rearrange $K$ directly.

  • If initial amounts are given instead, use an equilibrium table (Card 8).

• Step 5: Substitute back to verify that $Q = K$.

Useful relationships

• Gases: $p_i = c_iRT$, which links $K_p$ and $K_c$ (Card 6).

• Mole fraction and partial pressure: $p_A = x_A P_{total}$, where $x_A = n_A/n_{total}$ (as in your slide).

• Dalton’s law: $P_{total} = \sum p_i$.

Pitfalls

• Equilibrium values are not the initial values.

• Concentrations must be in mol L⁻¹, not moles, unless $\Delta n_{gas} = 0$ so that volume cancels.

• If you are solving for a pressure, use $K_p$ and pressures. If you are solving for a concentration, use $K_c$.

• Roots must be physically meaningful, meaning positive and not exceeding the available material.

• Taking a square root when the stoichiometric power is 2 is valid only for a single unknown with exponent 2.

Worked example
For $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$, $K_c = 4.6\times10^{-3}$ at 298 K. At equilibrium $[\text{N}_2\text{O}_4] = 0.100\ \text{M}$. Find $[\text{NO}_2]$.

1. $K_c = \dfrac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}$.

2. Rearrange: $[\text{NO}_2]^2 = K_c[\text{N}_2\text{O}_4] = (4.6\times10^{-3})(0.100\ \text{M})$. Since $K_c$ is dimensionless, the units are M².

   • $[\text{NO}_2]^2 = 4.6\times10^{-4}\ \text{M}^2$.

3. $[\text{NO}_2] = \sqrt{4.6\times10^{-4}} = \mathbf{0.0214\ M}$.

4. Check: $(0.0214)^2/0.100 = 4.6\times10^{-3}$. ✓

Predict the direction of a reaction, given K and the concentration of reactants and products.

Rule

• Compute $Q$, using the current concentrations or pressures in the same form as $K$, and compare it with $K$.

  • $Q < K$: too many reactants, so the reaction proceeds forward (toward products).

  • $Q = K$: the system is at equilibrium and there is no net change.

  • $Q > K$: too many products, so the reaction proceeds in reverse (toward reactants).

Why it works

• $\Delta G = RT\ln(Q/K)$, which follows from $\Delta G = \Delta G^\circ + RT\ln Q$ and $\Delta G^\circ = -RT\ln K$.

  • $Q < K$ gives a negative $\Delta G$ (forward spontaneous).

  • $Q > K$ gives a positive $\Delta G$ (reverse spontaneous).

• Molecularly, the forward and reverse rates are unequal away from equilibrium. The net flow continues until the rates match and $Q = K$.

Pitfalls

• $Q$ is not an equilibrium quantity. It can be evaluated for any composition, including the very start of a reaction.

• Do not compare raw concentrations. Always compare $Q$ with $K$.

• $K$ must be for the same equation, temperature, and form ($K_c$ versus $K_p$) as $Q$.

• Pure solids and liquids are omitted from $Q$ just as from $K$.

• If $K$ is huge, nearly every mixture has $Q < K$. If $K$ is tiny, nearly every mixture has $Q > K$.

• The direction tells you nothing about speed.

Worked example (continuing the slide’s NH₃ system)
$\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ at 298 K, $\Delta G^\circ = -33.32\ \text{kJ mol}^{-1}$, with $Q = 16/27 = 0.593$.

1. $\ln K = \dfrac{33320}{(8.314)(298)} = 13.45$, so $K = 6.9\times10^{5}$.

2. Compare: $Q = 0.593 \ll K = 6.9\times10^5$.

3. Since $Q < K$, the reaction proceeds forward, making more NH₃. This agrees with the negative $\Delta G$ from Card 2.

Convert between K and Kc.

Setup

• $K_c$ is built from molar concentrations. $K$ (the thermodynamic constant, also called $K_p$ when only gases are involved) is built from partial pressures.

• Both are dimensionless in their rigorous form.

Derivation

• For an ideal gas, $p_i = c_iRT$.

• Substitute into $K = \prod (p_i/p^\circ)^{\nu_i}$:

  • $K = \prod \left(\dfrac{c_iRT}{p^\circ}\right)^{\nu_i} = \prod \left(\dfrac{c_i}{c^\circ}\right)^{\nu_i}\left(\dfrac{c^\circ RT}{p^\circ}\right)^{\Delta n}$.

• With numerical values in bar and mol L⁻¹, this simplifies to:

  • $K = K_c\,(RT)^{\Delta n}$, or equivalently $K_c = K\,(RT)^{-\Delta n}$.

  • $\Delta n$ = (moles of gaseous products) − (moles of gaseous reactants).

  • $R = 0.083145\ \text{L bar mol}^{-1}\text{K}^{-1}$ when pressures are in bar.

  • $T$ is in K.

Pitfalls

• Count only gases in $\Delta n$. Solids, liquids, and dissolved species do not contribute.

• The value of $R$ must match the pressure unit. Using 8.314 J mol⁻¹K⁻¹ here is wrong. Use L·bar (0.08314) or L·atm (0.08206).

• If $\Delta n = 0$, then $K = K_c$.

• Using atm instead of bar changes $K$ slightly, because 1 atm = 1.01325 bar. The standard convention is bar.

• The conversion assumes ideal-gas behavior.

Worked example
$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ has $K_p = 0.113$ at 298.15 K. Find $K_c$.

1. $\Delta n = 2 - 1 = +1$.

2. $RT = (0.083145\ \text{L bar mol}^{-1}\text{K}^{-1})(298.15\ \text{K}) = 24.79\ \text{L bar mol}^{-1}$.

3. $K_c = K_p(RT)^{-\Delta n} = \dfrac{0.113}{24.79} = \mathbf{4.6\times10^{-3}}$.

4. This matches the $K_c$ used in Card 4.

Calculate the effect on K of reversing a reaction or multiplying the chemical equation by a factor.

Rules

• Reversing the reaction: $K_{rev} = 1/K$.

• Multiplying the equation by a factor $n$: $K_{new} = K^{\,n}$.

  • Halving the equation ($n = \tfrac12$) gives a square root.

• Adding reactions: $K_{total} = K_1 \cdot K_2 \cdots$.

Why: the connection to $\Delta G^\circ$ (which obeys Hess’s law)

• $\Delta G^\circ = -RT\ln K$.

• Reversing the equation flips the sign of $\Delta G^\circ$, so $\ln K \to -\ln K$ and $K \to 1/K$.

• Multiplying by $n$ multiplies $\Delta G^\circ$ by $n$, so $\ln K \to n\ln K$ and $K \to K^n$.

• Adding reactions adds their $\Delta G^\circ$ values, so the $\ln K$ values add and the $K$ values multiply.

Pitfalls

• $K$ is meaningless without its equation. Always state the balanced reaction it refers to.

• Multiplying the equation by 2 squares $K$. It does not double it.

• $Q$ transforms the same way. If $K$ is changed, $Q$ must be written for the same new equation.

• Do not add $K$ values when adding reactions. Multiply them.

• $K$ changes by these rules at fixed temperature. Temperature changes are a separate matter (Card 10).

Worked example (using the HI reaction from Card 3)
Given $\tfrac12\text{H}_2 + \tfrac12\text{I}_2(s) \rightarrow \text{HI}(g)$, $K = 0.50$.

• Reverse: $\text{HI} \rightarrow \tfrac12\text{H}_2 + \tfrac12\text{I}_2(s)$, $K = 1/0.50 = \mathbf{2.0}$.

• Double: $\text{H}_2 + \text{I}_2(s) \rightarrow 2\text{HI}$, $K = (0.50)^2 = \mathbf{0.25}$.

Combining with Hess’s law (your slide’s CO example)

• $\Delta G^\circ$ for $2\text{CO} + \text{O}_2 \rightarrow 2\text{CO}_2$ equals $+1088 + 2(-801) = -514$ kJ.

• In $K$ language, $K = (1/K_a)(K_b)^2$, where $K_a$ is for the first given reaction and $K_b$ for the second.

Use an equilibrium table to carry out equilibrium calculations.

The ICE method (Initial, Change, Equilibrium)

• Step 1: Write the balanced equation and the $K$ expression.

• Step 2: Initial row. Enter starting concentrations (or pressures). Use 0 for species not yet present.

• Step 3: Change row. Let the extent of reaction be $x$.

  • Each species changes by its coefficient times $x$, positive for products formed and negative for reactants consumed.

  • The direction is set by comparing $Q$ with $K$ (Card 5).

• Step 4: Equilibrium row. Add Initial and Change.

• Step 5: Substitute the equilibrium row into $K$ and solve for $x$.

• Step 6: Discard impossible roots (negative concentrations) and check by recomputing $K$.

Simplification (small-$x$ approximation)

• If $K$ is very small compared with the initial concentration, assume $x \ll$ initial amount and drop $x$ from sums such as $(0.50 - x) \approx 0.50$.

• Check: $x$ should be less than about 5% of the initial value. If not, solve the quadratic exactly.

Pitfalls

• The coefficients in the Change row must match the stoichiometry. For a $2\text{HI}$ product, the change is $+2x$, and in $K$ it is squared as $(2x)^2$.

• Work in concentrations (mol L⁻¹), not moles, unless $\Delta n_{gas} = 0$.

• If the volume changes, recompute the initial concentrations first.

• Solids and pure liquids do not get entries in $K$.

• With pressures, use $K_p$ and a pressure table. Do not mix the two.

Worked example
$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$, $K_c = 50.0$. Start with 1.00 M H₂ and 1.00 M I₂ and no HI.

	H₂	I₂	2HI
I	1.00	1.00	0
C	$$-x$$	$$-x$$	$$+2x$$
E	$$1.00 - x$$	$$1.00 - x$$	$$2x$$

1. $K_c = \dfrac{(2x)^2}{(1.00 - x)^2} = 50.0$.

2. The expression is a perfect square, so take the root: $\dfrac{2x}{1.00 - x} = \sqrt{50.0} = 7.071$.

3. Solve: $2x = 7.071 - 7.071x$, so $9.071x = 7.071$ and $x = 0.780$.

4. $[\text{HI}] = 2x = \mathbf{1.56\ M}$ and $[\text{H}_2] = [\text{I}_2] = 1.00 - 0.780 = \mathbf{0.220\ M}$.

5. Check: $(1.56)^2/(0.220)^2 = 50.3$, which is within rounding of 50.0. ✓

Use Le Chatelier’s principle to predict how the equilibrium composition of a reaction mixture is affected by adding or removing reagents, compressing or expanding the mixture, or changing the temperature.

Principle

• When a system at equilibrium is disturbed, it shifts to partially counteract the disturbance and re-establish equilibrium.

• It is a qualitative shortcut for the more fundamental test: compare $Q$ with $K$ after the disturbance.

1. Adding or removing a species (at constant $T$)

• $K$ does not change, but $Q$ does.

• Adding a reactant lowers $Q$ below $K$, so the system shifts forward.

• Adding a product raises $Q$ above $K$, so the system shifts in reverse.

• Removing a species shifts the system toward replacing it.

• Adding or removing a pure solid or liquid has no effect, because its activity is 1.

2. Compressing or expanding (gases)

• Compression increases pressure, and the system shifts toward the side with fewer moles of gas.

• Expansion shifts it toward the side with more moles of gas.

• If $\Delta n_{gas} = 0$, there is no shift.

• $K$ does not change.

• Molecular reason: concentrations of all gases scale together, but species raised to higher total powers in $Q$ respond more strongly.

• Adding an inert gas at constant volume changes no partial pressures, so there is no shift. Adding an inert gas at constant pressure increases the volume, which acts like expansion.

3. Changing temperature

• This is the only disturbance that changes $K$.

• Treat heat as a reagent:

  • Endothermic ($\Delta H^\circ > 0$): raising $T$ shifts the system forward and increases $K$.

  • Exothermic ($\Delta H^\circ < 0$): raising $T$ shifts the system in reverse and decreases $K$.

• The quantitative form is the van’t Hoff equation (Card 10).

4. Catalysts

• A catalyst speeds forward and reverse reactions equally. It changes neither $K$ nor the equilibrium composition, only the time to reach it.

Pitfalls

• Adding a reagent does not change $K$. It changes $Q$.

• Pressure changes caused by compression do not change $K$.

• Le Chatelier’s principle can mislead for complex combined disturbances. When in doubt, compute $Q$.

• “Shifts toward products” does not mean products exceed reactants. It means the equilibrium position moves in that direction.

Worked example
$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ at equilibrium. The volume is suddenly halved at constant $T$. Which way does it shift?

1. $K = \dfrac{(p_{\text{NO}_2}/p^\circ)^2}{p_{\text{N}_2\text{O}_4}/p^\circ}$.

2. Halving the volume doubles every partial pressure:

   • $Q = \dfrac{(2p_{\text{NO}_2})^2}{2p_{\text{N}_2\text{O}_4}} = 2\,\dfrac{p_{\text{NO}_2}^2}{p_{\text{N}_2\text{O}_4}} = 2K$.

3. $Q = 2K > K$, so the reaction shifts in reverse, toward $\text{N}_2\text{O}_4$.

4. This agrees with the rule: the reactant side has fewer gas moles (1 versus 2).

Predict the value of K at a different temperature.

The van’t Hoff equation

• From $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ and $\Delta G^\circ = -RT\ln K$:

  • $\ln K = -\dfrac{\Delta H^\circ}{RT} + \dfrac{\Delta S^\circ}{R}$.

• Writing this at two temperatures and subtracting gives $\ln\dfrac{K_2}{K_1} = -\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_2} - \dfrac{1}{T_1}\right)$.

  • $\Delta H^\circ$ is in J mol⁻¹.

  • $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$.

  • $T_1, T_2$ are in K.

Assumptions

• $\Delta H^\circ$ and $\Delta S^\circ$ are approximately independent of temperature over the range considered. This is a good approximation for modest temperature ranges.

• The plot of $\ln K$ versus $1/T$ is a straight line with slope $-\Delta H^\circ/R$.

Interpretation

• Endothermic ($\Delta H^\circ > 0$): the slope is negative, so $K$ increases as $T$ rises.

• Exothermic ($\Delta H^\circ < 0$): the slope is positive, so $K$ decreases as $T$ rises.

• Physically, the entropy term $T\Delta S^\circ$ grows in importance as $T$ rises, so entropy-driven processes become more favorable at high $T$.

Pitfalls

• Use kelvin and convert $\Delta H^\circ$ from kJ to J.

• The sign in front of $\Delta H^\circ/R$ is negative. Check your answer against the Le Chatelier direction in Card 9.

• Use $\Delta H^\circ$ for the equation as written.

• Your slide states that for a phase change $\Delta G^\circ = 0$. Strictly, $\Delta G = 0$ at equilibrium, and $\Delta G^\circ = 0$ only when $K = 1$, which for a vaporization means the normal boiling point at 1 bar.

Worked example (the slide’s Br₂ system)
$\text{Br}_2(l) \rightleftharpoons \text{Br}_2(g)$ has $\Delta H^\circ = 31\ \text{kJ mol}^{-1}$ and $\Delta S^\circ = 93\ \text{J K}^{-1}\text{mol}^{-1}$. At $T_1 = 333\ \text{K}$, $K_1 = 1$ (because $\Delta G^\circ = 0$ there). Find $K$ at $T_2 = 298.15\ \text{K}$.

1. $\dfrac{\Delta H^\circ}{R} = \dfrac{31000\ \text{J mol}^{-1}}{8.314\ \text{J mol}^{-1}\text{K}^{-1}} = 3729\ \text{K}$.

2. $\dfrac{1}{T_2} - \dfrac{1}{T_1} = \dfrac{1}{298.15} - \dfrac{1}{333} = 0.003354 - 0.003003 = 3.51\times10^{-4}\ \text{K}^{-1}$.

3. $\ln\dfrac{K_2}{K_1} = -(3729)(3.51\times10^{-4}) = -1.31$.

4. $K_2 = e^{-1.31} = \mathbf{0.27}$, since $K_1 = 1$.

5. Since $K = p_{\text{Br}_2}/p^\circ$, this predicts a vapor pressure of about 0.27 bar at 25 °C, lower than at 333 K, as expected for an endothermic process.

6. Cross-check: $\Delta G^\circ_{298} = 31000 - (298.15)(93) = 3272\ \text{J mol}^{-1}$, so $K = e^{-3272/2479} = 0.27$. ✓