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Approaching Equilibrium in the N2O4—NO2 System: A Common Mistake in Textbooks I. A. Leenson Department of Chemistry, Moscow State University, Moscow 119899, Russia;
[email protected] A clever demonstration with a dramatic or unpredictable effect always evokes students’ enthusiasm. An explanation of the effect observed should of course be not only apparent but also correct from the scientific point of view. Regrettably this is not always the case. One such example is the well-known demonstration of an equilibrium shift in a NO2–N2O4 gaseous mixture. Many journal articles describe the equilibrium in gas mixtures of nitrogen oxides both qualitatively and quantitatively, and several simple experiments have been developed that allow students to determine the equilibrium constants (1–10). This system is very convenient for such experiments because the equilibrium N2O 4(g) 2NO2(g) is readily shifted to the right or to the left upon changes in the temperature or volume (and therefore pressure) of the mixture. The observed vapor density of N2O4 at room temperature agrees with this dimeric formula, but it drops with rise of temperature until at 140 °C it indicates that only monomer NO2 is present (11). Of course, necessary precautions should be taken while conducting such experiments because nitrogen dioxide and tetroxide are highly toxic (12). Authors of many textbooks also describe an interesting experiment on the compression of NO2–N2O4 mixtures, or present questions for students concerning this experiment (13–21). Below is a typical description of the experiment (13). Some equilibria adjust themselves to new conditions so slowly that it is possible to watch the change under ordinary laboratory conditions, and thus see relaxation taking place. One such equilibrium is that between nitrogen dioxide (a dark brown gas) and its dimer dinitrogen tetroxide (a pale brown gas).1 … Quickly press in the plunger so as to compress the gas … and note the changes in color. It is best to hold the syringe in front a piece of white card while doing this. The color will first be seen to darken rapidly, as the concentration of NO2 increases, but will then slowly decrease, as the equilibrium adjusts to the new condition, higher pressure favoring the formation of N2O4.
Some textbooks present excellent pictures or photos of a transparent syringe with the gas mixture before and after compression. However, the explanation of color change is not always satisfactory. Accurate explanation should take into account two facts: noticeable heating upon compression and extremely fast chemical reactions in the system (relaxation time is of the order of a microsecond). The main goal of this paper is to present the accurate solution for the problem. The paper is suitable for an in-class lecture activity and will be useful for teachers and students in general chemistry and physical chemistry courses.
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Kinetics of Approaching Equilibrium in N2O4–NO2 System First of all it should be emphasized that the very high rate of both forward and reverse reactions makes it impossible to use any simple technique (such as a glass syringe) to study, even qualitatively, the kinetics of the processes in question. In fact, “the decomposition of N2O4 into NO2 is so rapid that it cannot be followed by classical kinetic methods” (22). As the authors of a recent paper in this Journal wrote, “Most relaxation times of interest are much shorter than mixing times, and the jump techniques were developed to overcome this by starting with systems already mixed and at equilibrium” (23). The rapid mixing and concentration jump experiments and other relaxation methods have been described in several papers; see for example refs 24–26. The reaction N2O4 2NO2 is extremely fast even at room temperature: the first-order rate constant k1 measured by supersonic absorption equals 1.7 × 105 s᎑1 at 25 °C, 2.4 × 105 s᎑1 at 30 °C, and 4.2 × 105 s᎑1 at 35 °C (27 ). Thus, at these temperatures decomposition of N2O4 proceeds with a halflife that is less than 10 microseconds! Moreover, the reverse reaction, the NO2 dimerization, is much faster. The measure of approaching equilibrium is given by the so-called characteristic (or relaxation) time τ; for the system A 2B, τ = 1/(k1 + 4k2[B]) where k1 and k2 are the rate constants for the dissociation and dimerization reactions, respectively (28). It is easy to estimate τ for this system. At room temperature (20 °C) the equilibrium constant Kp = P 2(NO2)/P(N2O4)
(1)
equals 0.10 (29). At atmospheric pressure when Pt = P(NO2) + P(N2O4)
(2)
is equal to 1 atm (Pt is the total pressure) we obtain from eq 1 P(NO2) ≈ 0.3 atm, then the second-order rate constant for the dimerization reaction, k2, is k2 = k1/Kp ≈ 106 atm s᎑1 and τ = 1/(k1 + 4k2P(NO2)) ≈ 10᎑6 s; that is, τ is of the order of one microsecond. This means that the equilibrium in the N2O4–NO2 mixture follows very strictly a piston that moves at any reasonable rate! In other words, the reaction proceeds too fast to be followed by a human eye. This result directly contradicts many explanations of the experiment, for example, as cited above, and especially the opinion expressed by a reader of this Journal in his letter to the editor (30): “Teachers need to focus student attention on the first few seconds of an NO2 dimerization system after it is disturbed and while a new equilibrium is becoming restored. Using ordinary means available to humans, such as a
Journal of Chemical Education • Vol. 77 No. 12 December 2000 • JChemEd.chem.wisc.edu
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rapid finger motion, it is possible to compress (or expand) the volume occupied by a mixture contained in a transparent syringe in which the equilibrium is established. Relative to the time span during which a new equilibrium is reached, several seconds, the compressing (or decompressing) action takes place in an instant.” It is obvious that “a rapid finger motion” cannot proceed on a microsecond time scale; and a new chemical equilibrium is not reached in “several seconds”, either. Thus we have to seek another explanation for the experimental facts described in the textbooks and Journal articles. Model Calculations of Mixtures at Equilibrium The authors of the textbooks attribute rapid darkening of the compressed gas to the decrease of the volume. This is correct but it is not the only reason. It is well known that many gases at ambient temperature are heated markedly upon compressing (everyone who has pumped a bicycle tire with a hand pump knows this very well). Any heating shifts the equilibrium markedly (according to eq 1, Kp being temperature dependent) because the N–N bond in N2O4 is very weak (less than 60 kJ/mol as compared to more than 300 kJ/mol for the rupture of the N–O bond in NO2). Thus the qualitative explanation of the experiment is rather simple. As the piston moves, the volume of the gas mixture is decreased and its temperature is increased. Both factors (especially heating) increase the concentration of NO2, and the mixture turns deep brown. Then heat is dissipated through the syringe walls (the rate of this process depends on the ambient temperature, on the thickness, and on the wall material). That is why it is impossible “to push down on the piston at constant temperature” as some textbooks propose (unless we are pushing down extremely slowly). Upon subsequent relatively slow cooling of a syringe (the thermal relaxation) the equilibrium is shifted to N2O4, and the mixture gradually fades and turns yellow-brown. How can we calculate the new equilibrium position just after a piston is pushed? If we know any three of the variables P, V, n, and T (n is the number of moles) that describe the physical state of a gas, we can calculate the fourth by using the ideal gas equation (31, 32). However, the quantitative calculations for the processes in question are not so simple because compressing the gas mixture changes not only its pressure, volume, and temperature but also the number of particles (or moles). Thus we cannot use explicitly any gas law. Below we present some model calculations, which are followed by the exact solution.
Initial Position of the Equilibrium For simplicity let the initial pressure of the gas mixture, P0, be 1 atm, the initial volume, V0, be 1 liter, the initial temperature, T0, be 25 °C, and the final volume be 0.5V0. For this and further calculations we shall need the equilibrium constant at different temperatures. There exist many polynomial equations for the temperature dependence of the equilibrium constant Kp for this system (see for example ref 4). But it is much simpler (with only a slight loss of precision) to use the thermodynamic equilibrium constant Kp = exp(᎑∆G °/RT ) = exp(21.13 – 6880/T )
(3)
The ∆G ° value for the reaction can be calculated from the thermochemical properties of NO2 (∆ fH ° = 33.18 kJ/mol,
∆S ° = 240.0 J mol᎑1 K᎑1) and of N2O4 (∆ fH ° = 9.16 kJ mol᎑1, ∆S ° = 304.2 J mol᎑1 K᎑1) at T = 298 K (33). For the dissociation reaction the standard change of enthalpy ∆Hr° is 57.2 kJ mol᎑1; the change of entropy ∆Sr° is 175.7 J mol᎑1 K᎑1. We can verify this equation as follows. Substitution of T = 293 K in eq 3 gives Kp = 0.095, in good agreement with the above cited experimental value of 0.10 at this temperature (29). Similarly, at T = 317 K, eq 3 gives Kp = 0.564—close to the 0.559 cited in ref 4. Thus we can use eq 3 as a very good approximation to calculate Kp at any temperature in the narrow temperature range. Now it is easy to calculate the initial gas composition at 25 °C. Equation 3 gives Kp = 0.141 at this temperature. From eqs 1 and 2 we have P(NO2) = [(Kp2 + 4KpPt)1/2 – Kp]/2 = 0.312 atm and P(N2O4) = 1.000 – 0.312 = 0.688 atm.
Adiabatic Compression of an Ideal Gas In all calculations we assume that nitrogen oxides behave as ideal gases. Of course it is only an approximation, but it is not a very bad one because the pressure is sufficiently low (near 1 atm). Thus, van der Waals’ constants, a and b, for NO2 and such gases as H2S and C2H6 are very similar (33): a ≈ 5 L2 atm, b ≈ 0.05 L mol᎑1, and calculations with these constants reveal a very slight (less than 1%) deviation from an ideal gas behavior for these gases when P ≈ 1 atm. An equation for the adiabatic reversible compression of an ideal gas is (34): γ
PV = const
(4)
where γ = Cp /Cv = 1.66 (monoatomic gas), 1.40 (diatomic gas), and 1.30 (triatomic gas) (32). Equation 4 can be transformed in order to calculate the final temperature after compression: γ
Tf /T0 = (V0 /Vf ) –1
(5)
where 0 and f represent initial and final states. If the volume of a syringe is adiabatically halved, eq 5 gives Tf /T0 = 20.30 = 1.23; and with T0 = 298 K, Tf = 298 × 1.23 = 367 K = 94 °C! For a polyatomic gas the temperature rise will be less dramatic. But it is obvious that any increase in temperature will inevitably induce a shift of the equilibrium to the right, favoring NO2 formation. Of course the real shift will not be so drastic because the lion’s share of the energy of a moving piston will be consumed by N–N bond rupture: the N–N bond energy is 57.2 kJ/mol as compared to heat capacities Cp = 77.3 J/(mol K) for N2O4 and Cp = 37.2 J mol᎑1 K᎑1 for NO2 (33). Nevertheless, the mixture will inevitably be heated: partial dissociation of N2O4 is impossible without heating. The real final temperature will be calculated in the last part of this section.2
Final (Thermal Relaxed) Position of the Equilibrium Here we shall examine the compressed mixture in thermal equilibrium with the environment (after cooling of the heated mixture). Let, as above, the initial volume, V0, be 1 liter and the final volume be 0.5V0, T = 298 K. The number of NO2 moles n0 (provided that only monomeric NO2 is present) is n0 = n(NO2) + 2n(N2O4) = const
(6)
Before compressing n(NO2) = PV/RT = 0.0128 mol and n(N2O4) = 0.0282 mol; n0 = 0.0128 + 0.0564 = 0.0692 mol.
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The total pressure Pt of the gas mixture is
1.1
Pt = P(NO2) + P(N2O4) = [n(NO2) + n(N2O4)]RT/V (7)
1.0 0.9
From eqs 1, 2, 6, and 7 we obtain (8)
and P(N2O4) = [4RTn0 /V – (Kp2 + 8RTKpn0/V )1/2 + Kp ]/8 (9) Equation 8 gives the partial pressure of NO2 at any temperature if n0 is known. After compression to 0.5 L and subsequent thermal relaxation to 298 K, P(NO2) = 0.456 atm (as compared with 0.312 atm before compressing). So the final (relaxed) gas mixture is 1.5 times darker than the initial state. The partial pressure of N2O4 after compression is P(N2O4) = 1.463 atm, and the total pressure is Pt = 1.92 atm. Note that this pressure is not twice as high as the initial pressure (1 atm), owing to the partial shift of the equilibrium toward N2O4 upon compression.
Adiabatic Compression of the Mixture of Nitrogen Oxides: Final (Thermally Unrelaxed) Position of the Equilibrium This problem cannot be solved using only usual ideal gas equations. It is necessary also to take into consideration the chemical reactions proceeding in this system. As a result, we must deduce a differential equation for the process and integrate it numerically. At infinitesimal movement of a piston work done by a piston (Pt dV ) is mainly consumed on the dissociation of an infinitesimal amount of N2O4 (the first term in the right part of eq 10) and partly on the heating of the gas mixture (the other 2 terms). Thus we obtain the equation PtdV = ∆Hr°dn(N2O4) + Cp(N2O4)n(N2O4)dT + Cp(NO2)n(NO2)dT
(10)
Taking into account that Pt = (n1 + n2)(RT /V ), at any moment we can rewrite eq 10 in a form convenient for numerical integration: (n1 + n2)(RT /V )(dV /dT ) = Hdn2/dT + (C1n1 + C2n2) (11) where n1 = n(NO2), n2 = n(N2O4), H = ∆Hr°, C1 = Cp(NO2), C2 = Cp(N2O4). From eqs 1 and 2 and from the ideal gas equations n1 = P(NO2)V /RT and n2 = P(N2O 4)V /RT, it follows that 2
V = n1 RT /(n 2 K p )
(12)
Equation 11 must be solved jointly with eqs 3, 6, and 12, because all the variables should adjust to each other upon compressing. Thus we have to solve the system of four equations: (n1 + n2)(RT /V )(dV /dT ) = Hdn2 /dT + (C1n1 + C2n2) n1 + 2n2 = n0 V = n12RT /(n2 K p ) Kp = exp(21.13 – 6880/T ) As the solution of these equations shows (see below), the temperature rises by only 13 K if the volume is halved, so we can safely neglect the slight temperature dependence of H, C1, and C2. Therefore, in these equations n1, n2, T, and V are variables and n0, H, C1, and C2 are constants.
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0.8
V/L
P(NO2 ) = [(K p2 + 8RTK pn0/V )1/2 – K p ]/4
0.7 0.9 0.5 0.4 0.3 0.2 295
300
305
310
315
320
325
T/K Figure 1. Variation of volume and temperature of the gas mixture (from 1 atm, 298 K) during fast adiabatic compression.
Figure 1 shows the calculated variation of volume and temperature of the mixture upon adiabatic compression.3 One can see that if volume is halved (from 1 to 0.5 L at initial pressure 1 atm) temperature rises to 310.7 K. It is easy to calculate the mixture composition at this temperature, as was done for 298 K. Equation 3 gives Kp = 0.363 at 310.7 K. From eqs 8 and 9 we have P(NO 2 ) = 0.719 atm and P(N2O 4) = 1.403 atm. Thus the observed darkening of the compressed mixture is rather profound (an increase in optical absorbance of the mixture is equal to 0.719/0.312 = 2.3), and the net reaction will shift strongly to NO2. The relaxation time to the initial temperature depends on the dimensions of a syringe, the heat conductivity of its walls, drafts in the laboratory, etc. Note that partial dimerization of NO2 upon cooling proceeds with the evolution of heat and slows down the lowering of temperature (as freezing water does in a closed space). In much the same way, partial dissociation of the dimer (as in the case of melting ice) prevents heating to high temperatures. It is useful to summarize all results obtained: Condition
≈T/K
P(NO2)/ atm
n(NO2)/ mol
Initial mixture
P(N2O4)/ n(N2O4)/ atm mol
≈298
0.312
0.0128
0.688
0.0282
Right after compression ≈311
0.719
0.0141
1.403
0.0275
≈298
0.456
0.00933
1.463
0.0299
The final state
where n0 = 0.0692 mol in all cases. Notes 1. It is an obvious mistake: nitrogen tetroxide is colorless. 2. After this paper had been completed I encountered (by chance) the paper written by J. A. Campbell, a prominent science teacher from UNESCO (35). Describing experiment No35 he writes: “Rapid compression of NO2 in a piston leads to intensification of color (due to adiabatic rise in T ) rather than the expected fading of color, which does follow shortly as T falls. (This effect is not due to a slow 2NO2 N2O4 equilibrium which is known to be very rapid from other data.)” Regretfully, this comment (hidden among 64 other experiments described by Campbell) obviously remained unnoticed.
Journal of Chemical Education • Vol. 77 No. 12 December 2000 • JChemEd.chem.wisc.edu
Information • Textbooks • Media • Resources 3. The solution for these equations was obtained through the use of Maple V3 program by M. I. Leenson, a graduate student of the Mathematical College, Independent University of Moscow.
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