3.10 Deep-Dive: NO2 Reaction
This section incorporates kinetics, equilibrium, and thermodynamics into an in-depth analysis of the following second-order gas phase reaction.
\[\mathrm{2NO_2}(g) \rightleftharpoons \mathrm{2NO}(g) + \mathrm{O_2}(g)\]
This reaction follows
- ΔH° > 0
- ΔS° > 0
This reaction is second-order and has been studied extensively. Thermodynamic and kinetic data reported in Evaluated Kinetic Data for High Temperature Reactions. Vol. 210) is given below.
See original pages (PDF).
We will analyze this reaction at various temperatures and characterize:
- Rate constant, k
- Rate law
- Integrated rate law
- Half-life, t1/2
- Thermodynamic characterizations (ΔH, ΔS, and ΔG)
- Equilibrium constants, Kp and Kc
- Equilibrium pressures/concentrations
3.10.1 Experimental Data
Thermodynamic Data
| T (K) |
ΔH° (kJ mol–1) |
ΔS° (J mol–1 K–1) |
log Kp (Kp in atm) |
log Kc (Kc in mol cm–3) |
|---|---|---|---|---|
| 298 | 114.28 | 146.35 | -12.40 | -16.79 |
| 300 | 114.31 | 146.45 | -12.26 | -16.66 |
| 500 | 116.29 | 151.70 | -4.23 | -8.84 |
| 1000 | 116.76 | 152.64 | 1.88 | -3.04 |
| 1500 | 115.73 | 151.81 | 3.90 | -1.19 |
| 2000 | 114.68 | 151.20 | 4.91 | -0.31 |
| 2500 | 113.99 | 150.89 | 5.51 | 0.19 |
| 3000 | 113.79 | 150.82 | 5.90 | 0.51 |
| 3500 | 114.08 | 150.90 | 6.19 | 0.73 |
| 4000 | 114.81 | 151.09 | 6.40 | 0.88 |
| 4500 | 115.92 | 151.35 | 6.57 | 1.00 |
| 5000 | 117.34 | 151.65 | 6.70 | 1.09 |
SI Units
\[\begin{align*} \log\left( K_{\mathrm{p}}/\mathrm{N~m^{-2}} \right ) &= \log\left ( K_{\mathrm{p}}/\mathrm{atm} \right ) + 5.006 \\[1.25ex] \log\left( K_{\mathrm{c}}/\mathrm{mol~m^{-3}} \right ) &= \log\left( K_{\mathrm{c}}/\mathrm{mol~cm^{-3}} \right ) + 6 \end{align*}\]
Recommended Rate Constant
\[\begin{align*} k &= 2.0\times 10^{12}~\mathrm{exp}\left({-13500/\mathrm{T}}\right)~\mathrm{cm^{3}~mol^{-1}~s^{-1}} \\[1.25ex] &= 3.3\times 10^{-12}~\mathrm{exp}\left({-13500/\mathrm{T}}\right)~\mathrm{cm^{3}}~\mathrm{molecule}^{-1}~\mathrm{s^{-1}} \end{align*}\]
(k is defined by -1/2d[NO2]/dt = k[NO2]2.)
Suggested Error Limits for Calculated Rate Constant: ±30% in temperature range 600–1000 K, but slightly greater at higher temperatures. Note: Expression is that of ROSSER and WISE (11).
Rate Parameters
\[\begin{align*} \log\left(\mathrm{A}/\mathrm{cm^3~mol^{-1}~s^{-1}}\right) &= 1230 \pm 0.07\\[1.25ex] \log\left(\mathrm{A}/\mathrm{cm^3~molecule^{-1}~s^{-1}}\right) &= -11.48 \pm 0.07\\[1.25ex] \mathrm{E_a/J~mol^{-1}} &= 112550 \pm 600 \\[1.25ex] \mathrm{E_a/cal~mol^{-1}} &= 26900 \pm 200 \end{align*}\]
A (cm3 mol–1 s–1) to A (L mol–1 s–1)
We generally use molar concentrations in mol L–1 in this class. Therefore, here we will convert the pre-exponential factor reported with units of cm3 (i.e. mL) into L.
\[\begin{align*} A &= 2.0\times 10^{12}~\mathrm{cm^{3}~mol^{-1}~s^{-1}} ~ \rightarrow \\ &= 2.0\times 10^{12}~\mathrm{mL~mol^{-1}~s^{-1}} \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \\ &= 2.0\times 10^{9}~\mathrm{L~mol^{-1}~s^{-1}} \rightarrow \\ &= 2.0\times 10^{9}~M^{-1}~\mathrm{s^{-1}} \end{align*}\]
Kp
The log of Kp can be determined from thermodynamic quantities.
\[\log K_p = \dfrac{-\left( \Delta H^{\circ} - T\Delta S^{\circ}\right)}{RT} \left ( \dfrac{1}{2.303} \right ) \]
or equivalently,
\[\log K_p = \dfrac{-\Delta G^{\circ}}{RT} \left ( \dfrac{1}{2.303} \right )\]
where R = 8.314 J mol–1 K–1.
Kp can then be determined using
\[K_{\mathrm{p}} = 10^{\log(K_{\mathrm{p}})}\]
Kp to Kc
The relation between Kp (in atm) and Kc (in mol L–1) is
\[K_{\mathrm{p}} = K_{\mathrm{c}} \left ( RT\right )^{\Delta n}\]
where Δn is the change in moles of gas (here, 1) and R = 0.08206 L atm mol–1 K–1.
Solving for Kc gives
\[K_{\mathrm{c}} = \dfrac{K_{\mathrm{p}}}{(RT)^{\Delta n}}\]
To reproduce the experimentally reported Kc given in mol cm–3, divide Kc (in mol L–1) by 1000.
log(Kp) to log(Kc)
The conversion from log(Kp / atm) to log(Kc / mol cm–3) is given as
\[\log\left (K_{\mathrm{c}} \right ) = \log\left(K_{\mathrm{p}}\right) - \Delta n\log(RT)\]
where Δn is the change in moles of gas (for this reaction, Δn is 1) and R = 82.05 cm3 atm mol–1 K–1 .
To convert log(Kc / mol cm–3) to log(Kc / mol L–1) use
\[\log\left (K_{\mathrm{c}}/\mathrm{mol~L^{-1}}\right ) = \log\left (K_{\mathrm{c}}/\mathrm{mol~cm^{-3}}\right ) + 3\] or, you can convert log(Kc / mol cm–3) to Kc (in mol cm–3) using
\[K_{\mathrm{c}} = 10^{\log\left( K_{\mathrm{c}}\right)}\]
followed by converting Kc (in mol cm–3) to Kc (in mol L–1) by using dimensional analysis and dividing by 1000 such that
\[K_{\mathrm{c}} \left(\mathrm{in}~ \frac{\mathrm{mol}}{\mathrm{cm}^{3}}\right ) \left ( \dfrac{1~\mathrm{cm^{3}}}{1~\mathrm{mL}}\right ) \left ( \dfrac{10^3~\mathrm{mL}}{1~\mathrm{L}}\right ) = K_{\mathrm{c}} \left( \mathrm{in}~\frac{\mathrm{mol}}{\mathrm{L}}\right )\]
and finally taking the log of Kc (in mol L–1).
3.10.2 Rate Constant
The rate constant for a reaction is temperature dependent and can be determined using the Arrhenius equation
\[k = Ae^{\frac{-E_{\mathrm{a}}}{RT}}\] where R = 8.314 J mol–1 K–1.
Rate Constant, k, at 298, 500, and 1000 K
Let us evaluate the rate constant for the given reaction at three different temperatures: 25 °C (298.15 K), 226.85 °C (500 K), and 726.85 °C (1000 K).
k at 298.15 K
\[\begin{align*} k &= Ae^{\frac{-E_{\mathrm{a}}}{RT}} \\ &= 2.0\times 10^{9}~M^{-1}~\mathrm{s^{-1}}~e^{\frac{-\left (112.55~\mathrm{kJ~mol^{-1}}\right )\left (\frac{10^3~\mathrm{J}}{\mathrm{kJ}}\right )}{\left (8.314~\mathrm{J~mol^{-1}~K^{-1}}\right )\left (298.15~\mathrm{K} \right)}} \\ &= 3.82\times 10^{-11}~M^{-1}~\mathrm{s^{-1}} \end{align*}\]
k at 500 K
\[\begin{align*} k &= Ae^{\frac{-E_{\mathrm{a}}}{RT}} \\ &= 2.0\times 10^{9}~M^{-1}~\mathrm{s^{-1}}~e^{\frac{-\left (112.55~\mathrm{kJ~mol^{-1}}\right )\left (\frac{10^3~\mathrm{J}}{\mathrm{kJ}}\right )}{\left (8.314~\mathrm{J~mol^{-1}~K^{-1}}\right )\left (500~\mathrm{K} \right)}} \\ &= 3.49\times 10^{-3}~M^{-1}~\mathrm{s^{-1}} \end{align*}\]
k at 1000 K
\[\begin{align*} k &= Ae^{\frac{-E_{\mathrm{a}}}{RT}} \\ &= 2.0\times 10^{9}~M^{-1}~\mathrm{s^{-1}}~e^{\frac{-\left (112.55~\mathrm{kJ~mol^{-1}}\right )\left (\frac{10^3~\mathrm{J}}{\mathrm{kJ}}\right )}{\left (8.314~\mathrm{J~mol^{-1}~K^{-1}}\right )\left (1000~\mathrm{K} \right)}} \\ &= 2.64\times 10^{3}~M^{-1}~\mathrm{s^{-1}} \end{align*}\]
Notice that the rate constant increases as temperature increases. The velocity of reaction is predicted to occur faster at higher temperatures.
Plots
Figure 3.28: Rate constant, k, vs. T
Figure 3.29: Arrhenius plot of ln(k) vs. 1/T
3.10.3 Rate Law
The reaction is experimentally determined to be second-order. Therefore, the rate law is
\[\mathrm{rate} = k[\mathrm{NO_2}]^2\]
Let us visualize the rate of reaction at 298, 600, and 1000 K if starting with some initial amount of NO2. Here we will assume [NO2]i = 1.0 M.
- Blue - Rate with k calculated at 298 K
- Red - Rate with k calculated at 500 K
- Black - Rate with k calculated at 1000 K
Figure 3.30: Rate vs. reactant concentration
As expected, the rate of reaction is much larger at higher temperature. This second-order reaction shows a large dependence of the rate on the concentration of reactants. As the reaction progresses and NO2 gets consumed, the rate of reaction slows down exponentially.
It can be difficult to see the rate plot at low temperature. We can take the natural log of the rate (i.e. ln(rate)) to better see these differences.
Figure 3.31: ln(rate) vs. reactant concentration
3.10.4 Integrated Rate Law
The integrated rate law for a second order reaction is given as
\[\dfrac{1}{[\mathrm{A}]_t} = kt + \dfrac{1}{[\mathrm{A}]_0}\]
We can use this to determine how long it will take to consume some initial amount of reactant. We will once again use the rate constant, k at three different temperatures and set our initial molar concentration of reactant to 1.0 M. Time is calculated at 0.1 M intervals.
- Blue - Time with k calculated at 298 K
- Red - Time with k calculated at 500 K
- Black - Time with k calculated at 1000 K
Figure 3.32: Reactant concentration vs. time
As expected, the time required to consume NO2 is much shorter at higher temperatures. As the reactant is consumed, the time of reaction increases since the rate of reaction is slowing down.
Given the difficult nature of comparing the data sets in the plot above, I plot the ln(time) vs. temperature relation below.
Figure 3.33: Reactant concentration vs ln(time)
We can “linearlize” the data by taking the inverse of the reactant concentration (as shown with the second-order integrated rate law). I only show the 298.15 K data for clarity. Notice how the data isn’t linear overall but across small temperature ranges, the data is close to linearity.
Figure 3.34: Inverse reactant concentration vs. time
3.10.5 Half-life
The half-life, t1/2, of a second-order reaction is given as
\[t_{1/2} = \dfrac{1}{k[\mathrm{A}]_0}\] and demonstrates that the half-life is dependent on the inverse of the reactant concentration; that is, as the reactant, A, gets consumed as the reaction progresses, the half-life decreases.
Half-life at 298, 500, and 1000 K
Let us evaluate the half-life, t1/2, of the reaction for the given reaction at three different temperatures: 25 °C (298.15 K), 326.85 °C (500 K), and 726.85 °C (1000 K). We will assume an initial concentration of 1.0 M for this problem.
t1/2 at 298.15 K
\[\begin{align*} t_{1/2} &= \dfrac{1}{k[\mathrm{A}]_0} \\ &= \dfrac{1}{3.82\times 10^{-11}~M^{-1}~\mathrm{s^{-1}}(1.0~M)} \\ &= 2.62\times 10^{10}~\mathrm{s}\\ &\approx 830~\mathrm{yr} \end{align*}\]
t1/2 at 500 K
\[\begin{align*} t_{1/2} &= \dfrac{1}{k[\mathrm{A}]_0} \\ &= \dfrac{1}{3.49\times 10^{-11}~M^{-3}~\mathrm{s^{-1}}(1.0~M)} \\ &= 2.87\times 10^{2}~\mathrm{s}\\ &\approx 4.77~\mathrm{min} \end{align*}\]
t1/2 at 1000 K
\[\begin{align*} t_{1/2} &= \dfrac{1}{k[\mathrm{A}]_0} \\ &= \dfrac{1}{2.64\times 10^{3}~M^{-1}~\mathrm{s^{-1}}(1.0~M)} \\ &= 3.79\times 10^{-4}~\mathrm{s}\\ &\approx 0.38~\mathrm{ms} \end{align*}\]
3.10.6 Free energy of reaction
The Gibbs free energy (ΔG°) of reaction can be calculated at various temperatures using the Gibbs free energy equation
\[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta{S}^{\circ}\]
The change in free energy allows us to predict if the reaction is thermodynamically favorable (i.e. spontaneous) or thermodynamically unfavorable (i.e. non-spontaneous). This assumes a closed system at a constant temperature.
| Sign | Spontaneity | Favor |
|---|---|---|
| ΔG < 0 | spontaneous | product favored |
| ΔG > 0 | non-spontaneous | reactant favored |
| ΔG = 0 | at equilibrium | none |
The experimental enthalpy of reaction (ΔH°) and entropy of reaction (ΔS°) are reported. However, we can predict with reasonable agreement these values by referring to the Standard Thermodynamic Values for Select Substances Table found in the appendix of this handbook.
Below are the tabulated standard thermodynamic values (at 25 °C) for our reaction.
| Substance | ΔHf° (kJ mol–1) |
S° (J mol–1 K–1) |
|---|---|---|
NO2(g) |
33.2 |
240.1 |
NO(g) |
90.25 |
210.8 |
O2 |
0 |
205.2 |
Use the appropriate enthalpy values in
\[\Delta H^{\circ} = \sum \nu \Delta H^{\circ} (\mathrm{products}) - \sum \nu \Delta H^{\circ} (\mathrm{reactants})\]
and entropy values in
\[\Delta S^{\circ} = \sum \nu S^{\circ} (\mathrm{products}) - \sum \nu S^{\circ} (\mathrm{reactants})\]
to predict the heat and entropy of reaction. Note that ν is the stoichiometric coefficient from the balanced chemical equation.
\[\begin{align*} \Delta H_{\mathrm{rxn}}^{\circ} &= \left [ (2 \times 90.25~\mathrm{kJ~mol^{-1}}) + (0~\mathrm{kJ~mol^{-1}}) \right ] - \left [ (2\times 33.2~\mathrm{kJ~mol^{-1}}) \right ] \\[1.25ex] &= 114.10~\mathrm{kJ~mol^{-1}} \\[2.00ex] \Delta S_{\mathrm{rxn}}^{\circ} &= \left [ (2 \times 0.2108~\mathrm{kJ~mol^{-1}~K^{-1}}) + (0.2052~\mathrm{kJ~mol^{-1}~K^{-1}}) \right ] - \left [ (2\times 0.2401~\mathrm{kJ~mol^{-1}~K^{-1}}) \right ] \\[1.25ex] &= 0.1466~\mathrm{kJ~mol^{-1}~K^{-1}} \end{align*}\]
These values are in close agreement with the experimentally reported values listed above. We see that the reaction is endothermic (ΔH° > 0). Furthermore, the entropy is positive (ΔS° > 0) which is expected since the reaction increases the moles of gas in the system (begins with 2 moles of gas and results in 3 moles of gas).
The remainder of this reaction exercise will use these values that we just computed.
Gibbs free energy, ΔG°, at 298, 500, and 1000 K
Let us evaluate the Gibbs free energy for the given reaction at three different temperatures: 25 °C (298.15 K), 326.85 °C (500 K), and 726.85 °C (1000 K).
ΔG° at 298.15 K
\[\begin{align*} \Delta G^{\circ} &= \Delta H^{\circ} - T\Delta{S}^{\circ} \\ &= 114.10~\mathrm{kJ~mol^{-1}} - (298.15~\mathrm{K})(0.1466~\mathrm{kJ~mol^{-1}~K^{-1}}) \\ &= 70.39~\mathrm{kJ~mol^{-1}} \end{align*}\]
ΔG° at 500 K
\[\begin{align*} \Delta G^{\circ} &= \Delta H^{\circ} - T\Delta{S}^{\circ} \\ &= 114.10~\mathrm{kJ~mol^{-1}} - (500~\mathrm{K})(0.1466~\mathrm{kJ~mol^{-1}~K^{-1}}) \\ &= 40.80~\mathrm{kJ~mol^{-1}} \end{align*}\]
ΔG° at 1000 K
\[\begin{align*} \Delta G^{\circ} &= \Delta H^{\circ} - T\Delta{S}^{\circ} \\ &= 114.10~\mathrm{kJ~mol^{-1}} - (298.15~\mathrm{K})(0.1466~\mathrm{kJ~mol^{-1}~K^{-1}}) \\ &= -32.5~\mathrm{kJ~mol^{-1}} \end{align*}\]
We see that at lower temperatures (298.15 and 500 K), the free energy of reaction is a positive value. At these temperatures, the reaction is thermodynamically unfavorable and the reaction will not proceed with vigor towards products. At the elevated temperature of 1000 K, the free energy of reaction is negative meaning the reaction is thermodynamically favorable. At this high temperature, we expect the reaction to proceed favorably towards products.
Plot
Figure 3.35: Gibbs free energy vs. temperature
Here we see that at low temperatures, ΔG°, is a positive value and the reaction is not thermodynamically favorable (exergonic). As the temperature increases, ΔG° becomes a negative value indicating that the reaction is thermodynamically favorable (exogonic) at higher temperatures. The temperature at which the reaction crosses over from being non-thermodynamically favorable to favorable can be determined by setting ΔG° = 0 and solving for T such that
\[\begin{align*} \Delta G^{\circ} &= \Delta H^{\circ} - T\Delta{S}^{\circ} \\[1.25ex] 0 &= \Delta H^{\circ} - T\Delta{S}^{\circ} \\[1.25ex] T &= \dfrac{\Delta H^{\circ}}{\Delta S^{\circ}} \\[1.25ex] &= \dfrac{114.1~\mathrm{kJ~mol^{-1}}}{0.1466~\mathrm{J~mol^{-1}~K^{-1}}} \\[1.25ex] &= 778~\mathrm{K} \end{align*}\]
3.10.7 Energy Diagram
We can create an energy diagram (or energy profile) for this reaction. Let us plot the Gibbs free energy of reaction as well as the activation energy of the transition state.
This energy diagram indicates that the reaction:
- occurs in a single step
- contains one transition state
- exergonic lower temperatures (thermodynamically unfavorable)
- exogonic at higher temperatures (thermodynamically favorable)
3.10.8 Equilibrium Constant, Kp
Since this is a gas phase reaction, we can express the equilibrium constant in terms of the pressures of the gas. Generally, we would only work with partial pressures of each gas for a gas phase reaction. Concentrations are more relevant for aqueous phase reactions.
We can determine the equilibrium constant, Kp, from the Gibbs free energy using
\[K_{\mathrm{p}} = e^{\frac{-\Delta G^{\circ}}{RT}}\]
where R = 8.314 J mol–1 K–1.
Equilibrium constant, Kp, at 298.15, 500, and 1000 K
Let us evaluate the equilibrium constant, Kp (in atm), for the given reaction using the appropriate Gibbs free energy (ΔG°) at three different temperatures: 25 °C (298.15 K), 226.85 °C (500 K), and 726.85 °C (1000 K).
Kp at 298.15 K
\[\begin{align*} K_{\mathrm{p}} &= e^{\frac{-\Delta G^{\circ}}{RT}} \\ &= e^{\frac{-\left (70.39~\mathrm{kJ~mol^{-1}}\right ) \left (\frac{10^3~\mathrm{J}}{\mathrm{kJ}}\right )}{\left (8.314~\mathrm{J~mol^{-1}~K^{-1}}\right ) \left (298.15~\mathrm{K} \right)}} \\ &= 4.65\times 10^{-13} \end{align*}\]
Kp at 500 K
\[\begin{align*} K_{\mathrm{p}} &= e^{\frac{-\Delta G^{\circ}}{RT}} \\ &= e^{\frac{-\left (26.14~\mathrm{kJ~mol^{-1}}\right ) \left (\frac{10^3~\mathrm{J}}{\mathrm{kJ}}\right )}{\left (8.314~\mathrm{J~mol^{-1}~K^{-1}}\right ) \left (500~\mathrm{K} \right)}} \\ &= 5.46\times 10^{-3} \end{align*}\]
Kp at 1000 K
\[\begin{align*} K_{\mathrm{p}} &= e^{\frac{-\Delta G^{\circ}}{RT}} \\ &= e^{\frac{-\left (-32.5~\mathrm{kJ~mol^{-1}}\right ) \left (\frac{10^3~\mathrm{J}}{\mathrm{kJ}}\right )}{\left (8.314~\mathrm{J~mol^{-1}~K^{-1}}\right ) \left (1000~\mathrm{K} \right)}} \\ &= 4.99\times 10^{1} \end{align*}\]
These values agree fairly well to the experimentally reported values. The orders of magnitude are the same.
Plot
Figure 3.36: Equilibrium constant (Kp) vs. temperature
3.10.9 Equilibrium Constant, Kc
The equilibrium constant, Kc, is determined from the concentrations of each species in the reaction. Concentration is generally reserved for aqueous phase reactions. However, we can interconvert between Kp and Kc using
\[K_{\mathrm{p}} = K_c(RT)^{\Delta n}\]
or, rearranged,
\[K_{\mathrm{c}} = \dfrac{K_{\mathrm{p}}}{(RT)^{\Delta n}}\]
where R = 0.08206 L atm mol–1 K–1 and Δn is the change in moles of gas for the reaction. Here, Δn is 1 since we start with 2 moles of gas (reactants) and end up with 3 moles of gas (products).
\[\begin{align*} \Delta n &= 3 - 2 \\ &= 1 \end{align*}\]
Equilibrium constant, Kc, at 298, 500, and 1000 K
Let us evaluate equilibrium constant, Kc (in mol L–1), for the given reaction using the corresponding Kp at three different temperatures : 25 °C (298.15 K), 226.85 °C (500 K), and 726.85 °C (1000 K).
I will use the Kp we have previously calculated instead of the experimentally reported values.
K at 298.15 K
\[\begin{align*} K_{\mathrm{c}} &= \dfrac{K_{\mathrm{p}}}{(RT)^{\Delta n}} \\ &= \dfrac{4.65\times 10^{-13}}{[(0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}})(298.15~\mathrm{K})]^{1}} \\ &= 1.68\times 10^{-14} \end{align*}\]
K at 500 K
\[\begin{align*} K_{\mathrm{c}} &= \dfrac{K_{\mathrm{p}}}{(RT)^{\Delta n}} \\ &= \dfrac{5.30\times 10^{-3}}{[(0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}})(500~\mathrm{K})]^{1}} \\ &= 5.46\times 10^{-5} \end{align*}\]
K at 1000 K
\[\begin{align*} K_{\mathrm{c}} &= \dfrac{K_{\mathrm{p}}}{(RT)^{\Delta n}} \\ &= \dfrac{49.85}{[(0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}})(1000~\mathrm{K})]^{1}} \\ &= 6.08\times 10^{-1} \end{align*}\]
Plot
Figure 3.37: Equilibrium constant (Kp) vs. temperature
3.10.10 Summary of Data
Below is a summary of all the values we have computed so far for our reaction of interest at three different temperatures.
| T (K) |
ΔH° (kJ mol–1) |
ΔS° (J mol–1 K–1) |
ΔG° (kJ mol–1) |
k (M–1 s–1) |
t1/2 (s) |
Kp (atm) |
Kc (mol L–1) |
|---|---|---|---|---|---|---|---|
| 298 | 114.1 | 146.6 | 70.39 | 3.82e-11 | 2.62e+10 | 4.65e-13 | 1.90e-14 |
| 500 | 114.1 | 146.6 | 40.80 | 3.49e-03 | 2.87e+02 | 5.46e-05 | 1.33e-06 |
| 1000 | 114.1 | 146.6 | -32.50 | 2.64e+03 | 3.79e-04 | 4.99e+01 | 6.08e-01 |
Calculated Data Using More Precise Thermodynamic Values
In our exercise of characterizing the reaction of interest, we have assumed that the enthalpy of reaction (ΔH) and the entropy of reaction (ΔS) does not change very much and has very little impact on ΔG (we applied values for ΔH and ΔS as measured at 25 °C to find ΔG). This is a common approximation to make. Note that if we were to use the appropriate ΔH and ΔS at 500 and 1000 K, respectively, the deviation from our approximate ΔG grows no more than 4.5 kJ mol–1 and has a small impact on our predicted equilibrium constants.
See the calculated values below when using ΔH and ΔS measured at 500 and 1000 K and compare the results to the table above. They are in close agreement thereby validating the approximation we used.
| T (K) |
ΔH° (kJ mol–1) |
ΔS° (J mol–1 K–1) |
ΔG° (kJ mol–1) |
k (M–1 s–1) |
t1/2 (s) |
Kp (atm) |
Kc (mol L–1) |
|---|---|---|---|---|---|---|---|
| 298 | 114.1 | 146.60 | 70.39 | 3.82e-11 | 2.62e+10 | 4.65e-13 | 1.90e-14 |
| 500 | 116.8 | 152.71 | 40.44 | 3.49e-03 | 2.87e+02 | 5.95e-05 | 1.45e-06 |
| 1000 | 116.9 | 152.88 | -35.98 | 2.64e+03 | 3.79e-04 | 7.58e+01 | 9.23e-01 |
To see how thermodynamic values change with temperature, see the NIST-JANAF Thermochemical Tables available at https://janaf.nist.gov/. I have plotted some of this data and included in the Appendix for the following substances:
I compare the data of ΔG° calculated with thermodynamic values measured at 25 °C (in blue) and ΔG° calculated with thermodynamic values measured at their corresponding temperatures (in dark red) below.
Figure 3.38: Gibbs free energy vs. temperature
Notice the relatively close agreement between the approximated ΔG° (blue) to the more precise ΔG° values (in dark red).
Here is a comparison of the equilibrium constants, Kp, as determined from ΔG° computed with thermodynamic data at 25 °C (in blue) and ΔG° values computed with thermodynamic data at their respective temperatures (in dark red).
Figure 3.39: Equilibrium constant, Kp, vs. temperature
Notice the relatively close agreement between the equilibrium constants determined using the approximated ΔG° (blue) vs. more precise ΔG° values (in dark red). At low temperatures, the agreement is better than at higher temperatures. However, at higher temperatures, even though the deviation increases, the equilibrium values have the same order of magnitude.
Plotting ln(Kp) vs temperature shows, visually, good agreement.
Figure 3.40: Equilibrium constant, ln(Kp), vs. temperature
Important Conclusions
- The rate constant k increases with increasing T (Arrhenius equation)
- The rate of reaction increases with increasing T (rate law)
- The half-life of reaction decreases with increasing T since the rate constant is larger at higher T
- The Gibbs free energy of reaction (ΔG°) decreases with increasing T
- The reaction is becoming more thermodynamically favorable at higher temperatures
- Note: This relation is not always true for all reactions. ΔG° would have to be reevaluated
- Example: A reaction that is exothermic (ΔH° < 0) whose entropy is increasing (ΔS° > 0) would always have a ΔG > 0 (it would always be thermodynamically favorable regardless of the temperature)
- The equilibrium constant, K, increases with decreasing ΔG°
- As the reaction becomes more thermodynamically favorable, the more products are favored
- At low temperature, reactants are favored and K is small
- At high temperature, products are favored and K is large
3.10.11 Equilibrium
The equilibrium constant represents a ratio of products to reactants when the reaction is at equilibrium. That is, at equilibrium the rate of the forward and reverse reaction are equal and the pressures (or concentrations) of all the substances in the mixture stop changing. We can determine the equilibrium pressures of a gas phase reaction (or concentrations of an aqueous phase reaction).
3.10.11.1 Equilibrium pressures at 298 K
The equilibrium constant, Kp, for our reaction of interest at 298.15 K (25 °C) is 4.65 × 10–13. Let us consider an initial NO2 pressure of 1 atm. We will determine the equilibrium pressures. Given the very small value for Kp, we would expect there to be predominantly reactant and very little product at equilibrium.
First, write out an ICE table.
- I means initial pressure or concentration
- C means change in pressure or concentration
- E means equilibrium pressure or concentration
Populate the I row with initial starting conditions (here we are only starting with 1 atm of NO2).
The C row is where we determine the change in pressure of each substance. Since the reaction is starting with only reactant, the reaction will proceed toward product thereby consuming NO2 and producing NO and O2. Take into account the stoichiometric relationships between all the substances from the balanced chemical equation.
Finally, obtain the E row by adding the first two rows together.
| 2NO2 | ⇌ | 2NO | + | O2 | |
|---|---|---|---|---|---|
I |
1.0 |
0 |
0 | ||
C |
-2x |
+2x |
+x | ||
E |
1.0-2x |
2x |
x |
We must now solve for x. Once we know the value for x, we can plug it into the expressions in the E row and determine the equilibrium pressures!
Write out the equilibrium expression.
\[\dfrac{(P_{\mathrm{NO}})^2(P_{\mathrm{O_2}})}{(P_{\mathrm{NO_2}})^2} = K_{\mathrm{p}}\]
Now we do some substitutions. Substitute the terms in the equilibrium row of the ICE table into the equilibrium expression. We will now also include the value for Kp (at 25°C).
\[\dfrac{(2x)^2(x)}{(1.0-2x)^2} = 4.65\times 10^{-13}\] We are now ready to solve for x.
Cubic Approach
You might notice that our equilibrium expression will lead to a cubic equation (there will be an x3, x2 and x term). These cannot be combined algebraically and therefore will require the evaluation of a cubic function to solve for x. Rearrange the equilibrium expression into the form of a cubic
\[\begin{align*} \dfrac{(2x)^2(x)}{(1.0-2x)^2} &= 4.65\times 10^{-13} \\ 4x^3 &= 4.65\times 10^{-13}(1.0-2x)^2 \\ 4x^3 &= 4.65\times 10^{-13}(1.0 - 4x + 4x^2) \\ 4x^3 &= 4.65\times 10^{-13} - 1.86\times 10^{-12}x + 1.86\times 10^{-12}x^2 \\ 4x^3 - 1.86\times 10^{-12}x^2 + 1.86\times 10^{-12}x - 4.65\times 10^{-13} &= 0 \end{align*}\]
We can solve for a cubic using WolframAlpha and typing in the equation!
Math string: 4x^3 - 1.86e-12x^2 + 1.86e-12x - 4.65e-13 = 0
The real solution to the equation is given as
\[x = 4.88\times 10^{-5}~\mathrm{atm}\]
Use the value for x to solve for the equilibrium pressures.
\[\begin{align*} P_{\mathrm{NO_2}} &= 1.0 - 2x \\ &= 1.0~\mathrm{atm} - 2(4.88\times 10^{-5}~\mathrm{atm}) \\ &\approx 1.0~\mathrm{atm} \\ P_{\mathrm{NO}} &= 2x \\ &= 2(4.88\times 10^{-5}~\mathrm{atm}) \\ &= 9.76\times 10^{-5}~\mathrm{atm} \\ P_{\mathrm{O_2}} &= x \\ &= 4.88\times 10^{-5}~\mathrm{atm} \end{align*}\]
As we concluded at the beginning of this problem, the mixture of substances at equilibrium (at 25 °C) is predominantly reactant and very little products.
Small x Approach
Given that we are starting the reaction out very close to equilibrium, we could potentially avoid having to solve a cubic function by pulling a cool math trick. We could try and assume that “x is small”. (Of course, we already know that x is a small number since we solved it already.)
Let’s take our equilibrium expression that we had written above.
\[\dfrac{(2x)^2(x)}{(1.0\color{red}{- 2x} )^2} = 4.65\times 10^{-13}\] Notice the “–2x” term highlighted in red. If we assume that x is a small number, then we can conclude that “two times a small number is still a small number” and “one minus a small number is essentially one”. See the math trick here?
We can rewrite the expression such that
\[\dfrac{(2x)^2(x)}{(1.0)^2} = 4.65\times 10^{-13}\] and now we can directly solve for x without running into a cubic function!
\[\begin{align*} \dfrac{(2x)^2(x)}{(1.0)^2} &= 4.65\times 10^{-13} \\ 4x^3 &= 4.65\times 10^{-13} \\ x^3 &= 1.16\times 10^{-13} \\ x &= 4.88\times 10^{-5} \end{align*}\]
Ah, much easier indeed!
Well, we invoked an approximation… the “small x approximation”. We need to test our hypothesis that x was “small” against some threshold or criteria. For this class, I will use the commonly quoted “5% rule”. Let me explain.
We originally stated that if x was a small number, then “1.0 - 2x” is approximately 1.0. Using the x we just solved for, I need to determine if “2x” is within 5% of the initial number, 1.0. (“Initial number” simply means the original number we were either adding x to or subtracting x from.)
\[\dfrac{2(4.88\times 10^{-5})}{1.0} \times 100\% = 9.8\times 10^{-3}~\%\]
We see that we fall well within the 5% threshold which means that our approximation worked well here and we can now solve for the equilibrium pressures using the x we obtained via the small x approximation.
\[\begin{align*} P_{\mathrm{NO_2}} &= 1.0 - 2x \\ &= 1.0~\mathrm{atm} - 2(4.88\times 10^{-5}~\mathrm{atm}) \\ &\approx 1.0~\mathrm{atm} \\ P_{\mathrm{NO}} &= 2x \\ &= 2(4.88\times 10^{-5}~\mathrm{atm}) \\ &= 9.76\times 10^{-5}~\mathrm{atm} \\ P_{\mathrm{O_2}} &= x \\ &= 4.88\times 10^{-5}~\mathrm{atm} \end{align*}\]
The equilibrium pressures determined here agree well with the equilibrium pressures determined when solving the cubic function.
3.10.11.2 Equilibrium pressures at 1000 K
The equilibrium constant, Kp, for our reaction of interest at 1000 K (726.85 °C) is 4.99 × 101. At this elevated temperature, the equilibrium constant is fairly large (greater than 1). Therefore, we would expect to have a predominance of product at equilibrium at this temperature.
We will follow the procedure used above.
| 2NO2 | ⇌ | 2NO | + | O2 | |
|---|---|---|---|---|---|
I |
1.0 |
0 |
0 | ||
C |
-2x |
+2x |
+x | ||
E |
1.0-2x |
2x |
x |
Write out the equilibrium expression followed by some substitutions.
\[\begin{align*} \dfrac{(P_{\mathrm{NO}})^2(P_{\mathrm{O_2}})}{(P_{\mathrm{NO_2}})^2} &= K_{\mathrm{p}} \\[1.25ex] \dfrac{(2x)^2(x)}{(1.0-2x)^2} &= 4.99\times 10^{1} \end{align*}\]
We note that our expression will result in a cubic equation.
Cubic Approach
Rearrange the equation in the form of a cubic.
\[\begin{align*} \dfrac{(2x)^2(x)}{(1.0-2x)^2} &= 4.99\times 10^{1} \\ 4x^3 &= 4.99\times 10^{1}(1.0-2x)^2 \\ 4x^3 &= 4.99\times 10^{1}(1.0 - 4x + 4x^2) \\ 4x^3 &= 4.99\times 10^{1} - 1.996\times 10^{2}x + 1.996\times 10^{2}x^2 \\ 4x^3 - 1.996\times 10^{2}x^2 + 1.996\times 10^{2}x - 4.99\times 10^{1} &= 0 \end{align*}\]
We can solve for a cubic using WolframAlpha and typing in the equation!
Math string: 4x^3 - 1.996e2x^2 + 1.996e2x - 4.99e1 = 0
Three solutions to the equation are provided.
\[\begin{align*} x_1 &= 0.456~\mathrm{atm} \\ x_2 &= 0.559~\mathrm{atm} \\ x_3 &= 48.88~\mathrm{atm} \\ \end{align*}\]
Can you rationalize which root to use? (Hint: Plug each x into the equilibrium term for NO2 and see if you can determine the correct root!)
Use the value for x to solve for the equilibrium pressures.
\[\begin{align*} P_{\mathrm{NO_2}} &= 1.0 - 2x \\ &= 1.0~\mathrm{atm} - 2(0.456~\mathrm{atm}) \\ &= 0.088~\mathrm{atm} \\ P_{\mathrm{NO}} &= 2x \\ &= 2(0.456~\mathrm{atm}) \\ &= 0.912~\mathrm{atm} \\ P_{\mathrm{O_2}} &= x \\ &= 0.456~\mathrm{atm} \end{align*}\]
As we concluded at the beginning of this problem, the mixture of substances at equilibrium (at 1000 K) contains much more products than reactants at room temperature.
Small x Approach
The small x approach should not work here since we are starting the reaction far away from equilibrium (x is expected to be large). I will show how it does not work below.
\[\begin{align*} \dfrac{(2x)^2(x)}{(1.0-2x)^2} &= 4.99\times 10^{1} \\ \dfrac{(2x)^2(x)}{(1.0)^2} &= 4.99\times 10^{1} \\ 4x^3 &= 4.99\times 10^{1} \\ x^3 &= 1.25\times 10^{1} \\ x &= 2.32 \end{align*}\]
If it is not immediate obvious that x is too large, we will go ahead and “test x” below.
\[\dfrac{2(2.32)}{1.0} \times 100\% = 464~\%\]
We see that we are well outside the 5% threshold which means that our approximation is a poor one and should not be used. Solving the cubic equation to obtain x is the only course of action.
Practice
What are the equilibrium pressures of the reaction when ΔG° = 0 when starting with [NO2]i = 1.0 M? ΔH° = 114.1 kJ mol–1 and ΔS° = 146.6 J mol–1 K–1