THE THERMAL DECOMPOSITION OF RIETHANE. I1 BY R. C. CANTELO
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The equation CH, C zH2 accurately expresses the methane equilibriuml. It is evident from this that the rate of decomposition of methane is proportional to the concentration of methane at any instant; while the reverse reaction proceeds at a rate proportional to the square of the concentration of hydrogen. In the following treatment x = decrease in partial pressure of methane a t time t . x, = decrease in partial pressure of methane when t = a i.e. when equilibrium has been reached. The following table is then self-explanatory.
TABLEI Partial Pressures CHI C 2Hz a b a-x b+zx a - x, b PX,
Time
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Then d(b zx)/dt = kl(a - x) - kz(b ZX)~ z dx/dt = ki(a - x) - kz(b At equilibrium o = kl(a - x,) - kz(b 2 ~ ~ ) ~ 4kz[b(x, - x) (x2, - xz) ] Subtracting z dx/dt = kl(x, - x) = (x, - X) [ki 4k2(b xa X) ] = 4k2(Xor X) [k1/4kz b x a X] But ki/kz = IL = P ~ H ~ / P C H ~ .*. dx/dt = zkz(x, - X) [I(/4 b XK X] Put [E/4 b x,] = B. Then dx/dt = 2k2(x, - x) (x B). dx/(x, - X) (X B) = zkzdt which integrated and the constant of integration added gives or In (x B)/(x, x) - In B/x, = zkz(x, B)t (X B)/(xa x,) log B/x, = zkz(xm B)t X 0.4343 log B)t 0%. 1) log (x B)/(x, - x) - log B/x, = 0.8686 kz(x, from which kz can be determined for any value of t, when x and x, are known. The meaning of x and of x, requires careful consideration. These are changes in the partial pressure of methane, respectively a t time t and when equilibrium has been reached. From the equation CH, C zH2, it is evident that when the partial pressure of methane has decreased by x the total pressure has become I x. Thus x and x, are not values measured under atmospheric pressure but under a pressure of I x atmospheres. It is usual, however, in experimental work
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'Bone and Coward: J. Chem. SOC., 93, 1197 (1908); Mayer and Altmayer: Ber. 40, 2134 (1907); Coward and Wilson: J. Chem. SOC., 115, 1380 (1919).
900
R. C. CANTELO
to allow the volume to change while the pressure is kept constant at one atmosphere. The partial pressures obtained under these conditions are obviously such that the total pressure is one atmosphere, Yet the values of x and x, can be calculated from analytical data for atmospheric pressure. The “physical” correction for the partial pressure of hydrogen determined under a total pressure of one atmosphere is given by the equation 2x, / ( I +x,) = pH2, but in addition it is necessary strictly to apply a “chemical” correction as well. This will be apparent from the following concrete example. For IOOO~C., the equilibrium concentration of hydrogen, calculated by the method given below, is 0.98 atmospheres. 2xcc/(1 x, = 0.98, whence xcc = 0.96, i.e. the partial pressures of hydrogen and methane respectively are 1.92 and 0.04 atmospheres under a total pressure of 1.96 atmospheres, But P ~ H J P C H ~= K,; and it is evident that the value of the expression on the left hand side of the equation tends to increase, assuming no reaction to take place. Therefore, to restore the equilibrium ratio K,, combination of hydrogen with carbon must occur to form methane. In the present example, let y = this decrease in partial pressure of hydrogen. Then (1.92 - YI2 = K, = 5 5 . 5 (for 1000°C.) Lo4 Y) y by trial is found to be between 0 . 0 2 and 0.03. Since, however, the value 0.98 calculated for the equilibrium concentration of hydrogen is itself in error 3y0 a t least, the refinement obtained by applying this second correction (in itself strictly accurate) would be solely imaginary. The writer has therefore used the expression 2xo:/(I - xcc) = PH1 as a first approximation to determine x,. I n the case of the determination of x, the “physical” correction is the only one that can be applied. It is possible now to test Equation I by data for 985OC., given by Bone and Coward1. Table I1 gives the results of a series of experiments in which methane was shut up in heated tubes (unpacked) at atmospheric pressure for different periods of time.
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TABLE I1 Temperature 985 985 IO00 985 985 Time in minutes 30 60 I 5 I5 nil nil 0.5 nil C2Hz 0.5 0.3 nil C& 0.3 1.3 0.5 61.85 48.2 65.25 CH4 90.4 75.4 38.8 51.2 33 * 3 Hz 8.8 23.6 For purposes of calculation we can call the temperature I O O O O C . , and it is necessary now to determine K, and xa for this temperature. Saunders2 develops the following equation for the equilibrium constant for the reaction between methane, hydrogen and amorphous carbon : J. Chem. SOC., 93, 1197 (1908). J. Phys. Chem., 28, 1151 (1924).
901
THERMAL DECOMPOSITION O F METHANE,
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log Kp = 4,583/T - 1.75 log T 0.000630 T - 0.7 and shows that this gives results in agreement with Mayer and Altmayer's' determined constants, For T = 1273'K. this gives ,Kp = 55.5, from which the equilibrium values of hydrogen and methane can be obtained by the method given by the writer in a previous paper2, pH2 = 0.98, whence x, = 0.96. Again in this series of experiments b = 0 , so that B = 55.5/4 .96 = 14.86. It is necessary now to calculate the values of x corresponding to the analytical data of Table 11. These values follow: I 5 I5 30 60 Time (mins.) X 0.046 0 . I34 0.200 0.241 0.345 The fact that x a t the end of 60 minutes is 0,345 while xcc = 0.96 is in itself sufficient evidence of the slowness of the decomposition of methane. Equation I may be rearranged: I x + B I3 k2 = 'log - - log 0.8686 (x, B)t X, -x XCC Table I11 gives the calculated values of kz.
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TABLE 111 Time (mins.) I 5 I5 30 60 k2 X 103 1.7 0.96 0.52 0.32 0.32 Bone and Coward3 emphasize the fact that an abnormal period occurs at the start of the reaction, this in their opinion being due to the reaction taking place a t the walls of the reaction tube. After these have become coated with deposited carbon, the reaction proceeds to a greater degree normally. Table IV gives the values for k2 X 103 calculated with the fifth minute considered as the zero point. TABLE IV Time (mins.) I 5 15 30 60 kz X 103 0.24 0.16 0.15 The agreement is much better. To complete the mathematical analysis, the graphical solution js added : Y = log x B/x, - x Let b = log B/x, B) m = 0.8686 kz(x, Then Y =mt b and if the values of Y as ordinates be plotted against the values of t as abscissae, the points will lie on a straight line. The most probable straight line may be drawn by determining two Average Points and connecting them. The slope of this line gives m; from which k, (the value of kz for the whole process) may be calculated.
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Department of Chemistry, Unizwsity of Cincinnati, Cincinnati, Ohio. Ber., 40, 2134 (1907). Csntelo: J. Phys. Chem., 28, 1035 (1924) 3 J. Chem. SOC.,93, 1197 (1908).
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