Chem 351 Help — Gas-Phase Equilibrium

Gas-Phase Equilibria: From Equilibrium Composition to Thermodynamic Quantities

For a gas-phase reaction, the equilibrium composition contains a tremendous amount of thermodynamic information. By measuring the equilibrium partial pressures of the reactants and products, we can determine the equilibrium constant and ultimately the standard Gibbs energy change for the reaction.

Consider the general reaction

\[ aA(g)+bB(g) \rightleftharpoons cC(g)+dD(g) \]

The equilibrium constant expressed in terms of partial pressures is

\[ K_p = \frac{p_C^c p_D^d} {p_A^a p_B^b} \]

Once the equilibrium partial pressures are known, the value of \(K_p\) can be calculated directly.

Relating \(K_p\) and \(K_c\)

Equilibrium constants may also be expressed in terms of concentrations:

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

For gas-phase reactions, \(K_p\) and \(K_c\) are related through the ideal gas law:

\[ K_p = K_c(RT)^{\Delta n} \]

where

\[ \Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]

Thus, once \(K_p\) is known, \(K_c\) can be calculated directly.

Relating the Equilibrium Constant to \(\Delta G^\circ\)

The equilibrium constant is connected to thermodynamics through the relationship

\[ \Delta G^\circ = -RT\ln K \]

For ideal gases, \(K\) is well approximated by \(K_p\), allowing the standard Gibbs energy change to be determined from equilibrium measurements.

The sign and magnitude of \(\Delta G^\circ\) provide information about the position of equilibrium:

\(\Delta G^\circ < 0\): products are favored.
\(\Delta G^\circ = 0\): neither side is favored.
\(\Delta G^\circ > 0\): reactants are favored.

Strategy for Solving Gas-Phase Equilibrium Problems

Use the equilibrium composition (often obtained from an ICE table) to determine the equilibrium partial pressures.
Substitute the equilibrium partial pressures into the expression for \(K_p\).
Convert \(K_p\) to \(K_c\) if needed using \[ K_p=K_c(RT)^{\Delta n} \]
Calculate \(\Delta G^\circ\) using \[ \Delta G^\circ=-RT\ln K \]

Big picture: Measurements of equilibrium composition allow us to determine equilibrium constants, and equilibrium constants allow us to determine thermodynamic quantities such as \(\Delta G^\circ\). In this way, equilibrium measurements provide a direct connection between experimental observations and the thermodynamic driving forces behind chemical reactions.

Worked examples

Worked Example: Gas-Phase Equilibrium from Partial Pressures

Consider the gas-phase reaction at 298 K:

\[ A(g)+3B(g)\rightleftharpoons2C(g) \]

A 1.00 L fixed-volume container is charged with \(1.00\ \text{atm}\) of \(A\) and \(2.00\ \text{atm}\) of \(B\). At equilibrium, the partial pressure of \(C\) is found to be \(0.60\ \text{atm}\). Determine \(K_p\), \(K_c\), and \(\Delta G^\circ\).

Solution

	\(A\)	\(B\)	\(C\)
Initial	1.00 atm	2.00 atm	0
Change	\(-x\)	\(-3x\)	\(+2x\)
Equilibrium	1.00 atm - x	2.00 atm - 3x	2x

Since \(2C\) is formed, let the reaction progress correspond to \(x\):

\[ 2x=0.60\ \text{atm} \] \[ x=0.30\ \text{atm} \]

As such,

	A	B	C
Equilibrium	0.70 atm	1.10 atm	0.60 atm

Therefore,

\[ K_p= \frac{p_C^2}{p_Ap_B^3} \] \[ K_p= \frac{(0.60\ atm)^2}{(0.70\ atm)(1.10\ atm)^3} \] \[ K_p=0.387\ atm^{-2} \]

For this reaction,

\[ \Delta n = 2-(1+3)=-2 \]

so

\[ K_p=K_c(RT)^{-2} \] \[ K_c=K_p(RT)^2 \] \[ K_c=(0.387\ atm^{-2})[(0.08206 \frac{atm\ L}{mol\ K})(298\ K)]^2 \] \[ K_c=2.31\times10^2\ L^2\ mol^{-2} \]

Finally,

\[ \Delta G^\circ=-RT\ln K_p \] \[ \Delta G^\circ = -(8.314\frac{J}{mol\ K})(298\ K)\ln(0.387) \] \[ \Delta G^\circ=2.35\times10^3\ \text{J mol}^{-1} \] \[ \boxed{\Delta G^\circ=2.35\ \text{kJ mol}^{-1}} \]

Physical interpretation: Since \(K_p<1\), the equilibrium is reactant-favored under standard-state conditions, and \(\Delta G^\circ\) is positive. However, the reaction still forms a measurable amount of product because the actual equilibrium composition depends on the starting partial pressures.

Practice

Problem 1
Consider the reaction \[ A(g)+2B(g)\rightleftharpoons C(g) \] A 1.00 L container is charged with \[ p_A=1.00\ \text{atm} \] and \[ p_B=2.00\ \text{atm} \] Initially, no \(C\) is present. At equilibrium, \[ p_C=0.50\ \text{atm} \] Calculate: \(K_p\) \(K_c\) \(\Delta G^\circ\) at 298 K
Show Answer Since \[ p_C=x=0.50\ \text{atm} \] the equilibrium pressures are \[ p_A=0.50\ \text{atm} \] \[ p_B=1.00\ \text{atm} \] \[ p_C=0.50\ \text{atm} \] Therefore, \[ K_p = \frac{0.50} {(0.50)(1.00)^2} = 1.00 \] Since \[ \Delta n=1-(1+2)=-2 \] \[ K_c = K_p(RT)^2 \] \[ K_c = (1.00)(24.45)^2 = 5.98\times10^2 \] Finally, \[ \Delta G^\circ = -RT\ln K_p = 0 \] \[ \boxed{\Delta G^\circ=0} \]

Problem 2
Consider the reaction \[ A(g)+B(g)\rightleftharpoons2C(g) \] A reaction vessel is initially charged with \[ p_A=1.50\ \text{atm} \] and \[ p_B=1.50\ \text{atm} \] with no product present. At equilibrium, \[ p_C=0.80\ \text{atm} \] Calculate: \(K_p\) \(K_c\) \(\Delta G^\circ\) at 298 K
Show Answer Since \[ 2x=0.80 \] \[ x=0.40 \] Equilibrium pressures: \[ p_A=1.10\ \text{atm} \] \[ p_B=1.10\ \text{atm} \] \[ p_C=0.80\ \text{atm} \] Therefore, \[ K_p = \frac{(0.80)^2} {(1.10)(1.10)} \] \[ K_p=0.529 \] Since \[ \Delta n=2-(1+1)=0 \] \[ K_c=K_p \] \[ \boxed{K_c=0.529} \] Finally, \[ \Delta G^\circ = -(8.314)(298)\ln(0.529) \] \[ \boxed{\Delta G^\circ=1.58\ \text{kJ mol}^{-1}} \]

Problem 3
Consider the reaction \[ 2A(g)\rightleftharpoons B(g) \] A container is charged with \[ p_A=2.00\ \text{atm} \] and no product. At equilibrium, \[ p_B=0.40\ \text{atm} \] Calculate: \(K_p\) \(K_c\) \(\Delta G^\circ\) at 298 K
Show Answer Since \[ x=0.40\ \text{atm} \] and \[ 2A\rightarrow B \] the equilibrium pressure of A is \[ p_A = 2.00-2(0.40) = 1.20\ \text{atm} \] Therefore, \[ K_p = \frac{0.40} {(1.20)^2} \] \[ K_p=0.278 \] Since \[ \Delta n=1-2=-1 \] \[ K_c = K_p(RT) \] \[ K_c = (0.278)(24.45) \] \[ \boxed{K_c=6.80} \] Finally, \[ \Delta G^\circ = -(8.314)(298)\ln(0.278) \] \[ \boxed{\Delta G^\circ=3.18\ \text{kJ mol}^{-1}} \]

Problem 3

Consider the reaction \[ 2A(g)\rightleftharpoons B(g) \] A container is charged with \[ p_A=2.00\ \text{atm} \] and no product. At equilibrium, \[ p_B=0.40\ \text{atm} \] Calculate:

\(K_p\)
\(K_c\)
\(\Delta G^\circ\) at 298 K

Show Answer

Since \[ x=0.40\ \text{atm} \] and \[ 2A\rightarrow B \] the equilibrium pressure of A is \[ p_A = 2.00-2(0.40) = 1.20\ \text{atm} \] Therefore, \[ K_p = \frac{0.40} {(1.20)^2} \] \[ K_p=0.278 \] Since \[ \Delta n=1-2=-1 \] \[ K_c = K_p(RT) \] \[ K_c = (0.278)(24.45) \] \[ \boxed{K_c=6.80} \] Finally, \[ \Delta G^\circ = -(8.314)(298)\ln(0.278) \] \[ \boxed{\Delta G^\circ=3.18\ \text{kJ mol}^{-1}} \]

Key points (one glance)

The equilibrium composition of a gas-phase reaction can be used to determine the equilibrium constant and important thermodynamic quantities.
Equilibrium partial pressures are often obtained using an ICE table and the measured equilibrium pressure of one species.
For a general gas-phase reaction \[ aA+bB\rightleftharpoons cC+dD \] the equilibrium constant expressed in terms of partial pressures is \[ K_p = \frac{p_C^c p_D^d} {p_A^a p_B^b} \]
Once the equilibrium partial pressures are known, \(K_p\) can be calculated directly.
For gas-phase reactions, \[ K_p = K_c(RT)^{\Delta n} \] where \[ \Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
If \[ \Delta n=0 \] then \[ K_p=K_c \]
The equilibrium constant is related to the standard Gibbs energy change through \[ \Delta G^\circ = -RT\ln K \]
The sign of \(\Delta G^\circ\) indicates the position of equilibrium:
- \(\Delta G^\circ<0 \Rightarrow K>1\)
- \(\Delta G^\circ=0 \Rightarrow K=1\)
- \(\Delta G^\circ>0 \Rightarrow K<1\)
A large equilibrium constant indicates that products are favored at equilibrium, while a small equilibrium constant indicates that reactants are favored.
The equilibrium constant describes the equilibrium composition of a system, while \(\Delta G^\circ\) provides the thermodynamic interpretation of that equilibrium.

Big picture: Measurements of equilibrium composition allow us to determine equilibrium constants, and equilibrium constants allow us to determine thermodynamic quantities such as \(\Delta G^\circ\). This connection provides a direct bridge between experimental equilibrium measurements and the thermodynamic driving forces that govern chemical reactions.