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Chemical Equilibrium, Law Of Mass-action : Law Of Mass-action, Kinetic Deduction |
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We may consider the law of mass-action as an experimental fact. It may, however, be deduced: (1) from thermodynamics; (2) from the kinetic theory. A sketch of the second method, due to Guldberg and Waage, will be given here. Consider the formation of hydrogen iodide from hydrogen and gaseous iodine. Molecules of HI can be formed only as the result of collisions of iodine and hydrogen molecules, the number of collisions per second being proportional to the number of molecules of each gas present in unit volume, i.e., to its concentration. It is therefore proportional to the product of these concentrations, k [H2] x [I2]. Every collision may not result in the formation of hydrogen iodide, but a definite fraction x of the total number of collisions will be effective; hence the rate of formation of HI is equal to xk [H2] [I2], or k1 [H2] [I2] where k1 = xk, and x, k, are constants. Similarly, the speed of decomposition of HI will be k2[HI]2, since two HI molecules must collide, and the probability for this is proportional to [HI]2. The two reactions: (a) H2 + I2 => 2HI, (b) 2HI => H2 + I2, go on simultaneously; hence: Rate of formation of HI = Rate of combination of H2 and I2 to HI - Rate of decomposition of HI = k1[H2] x [I2] - k2[HI]2. This may be positive, negative, or zero, according to the values of k1[H2] x [I2] and k2 [HI]2. When the rate of formation of HI is zero, the system is in equilibrium since then HI is decomposed exactly as fast as it is formed, so that the amount of HI is independent of the time. Hence in equilibrium: k1[H2] x [I2] - k2[HI]2 = 0; k1[H2][I2] = k2[HI]2 or  At a given temperature, K is constant: it is the equilibrium constant. It is independent of the amounts of iodine, hydrogen, and hydrogen iodide originally taken, but depends on the temperature. Those molecules which are in a condition to undergo chemical change on collision (active molecules) appear to be those possessing more than a certain critical amount of energy. When a molecule acquires this critical increment of energy it becomes active. The number of bimolecular collisions in a mixture of hydrogen and iodine vapour can be calculated from the kinetic theory. At room temperature at atmospheric pressure this number is readily found to be of the order of 108; it is, further, proportional to √T. Experiments show, however, that the rate of chemical change is much smaller than would be expected if every collision were effective and also that it increases with rise of temperature much faster than √T. In fact, the velocity of most chemical reactions is approximately doubled for a rise of 10° C. In order to explain such results, it is assumed that reaction on collision occurs only when the colliding molecules are activated, or possess energies above the average value for the temperature of the gas. The effect of temperature is then explained if it is assumed that the proportion of active molecules increases with temperature according to an exponential factor e-q/kT. where k is Boltzmann's constant and q is the energy of activation; per mol Q = N0q. At 556° abs. the number of collisions in HI at a concentration of 1 mol per litre is 6 x 1034. The measured rate of decomposition into H2 and I2 is 2 x 1017 molecules per second;  = 2x1017/6x1034 = e-Q/RT;2.3026 x log10(2x1017/6x1034) = -Q/2 x 556; Q = 47.420 g. cal. Example. - 7.94 c.c. of hydrogen (at S.T.P.) and 0.0601 gm. of solid iodine were heated in a sealed bulb at 444° until equilibrium was reached. 9.52 c.c. of hydrogen iodide (at S.T.P.) were formed. Now at S.T.P. 2 x 127 gm. of iodine (I2) occupy 22,420 c.c.
vol. of I2 vapour at S.T.P. initially present = 22420 x 0.0601 / (2 x 127) = 5.30 c.c. The 9.52 c.c. of HI are formed from 4.76 c.c. of H2 and 4.76 c.c. of I2. in equilibrium: vol. of H2 = 7.94 - 4.76 = 3.18 c.c. (4.76 = 0.5 x vol. of HI = 0.5 x 9.52). vol. of I2 = 5.30 - 4.76 = 0.54 c.c. vol. of HI = 9.52. Hence, if V is the volume of the bulb in litres, the concentrations are: [H2] = 3.18/224207; [I2] = 0.54/22420V; [HI] = 9.52/22420V. Hence:  = 3.18 x 0.54 / (9.52)2 = 0.01895.Now, suppose 8.10 c.c. of hydrogen and 2.94 c.c. of iodine vapour (at S.T.P.) heated at 444°. What volume of HI will be formed in equilibrium? Let 2x c.c. be formed : Volumes: H2(8.10 -x) + I2(2.94-x) <=> 2HI(2x) (8.10 -x)(2.94-x)/4x2 = 0.01895; x = 2.825 or 9.12. Only the root 2.83 is admissible, since 2.94 c.c of I2 vapour can give only 5.88 c.c. of HI as a maximum. Thus, the volume of HI formed = 2 x 2.83 c.c. = 5.63 c.c. Bodenstein by experiment found 5.66 c.c. |
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