These are plotted on figs. 2 and 3, and the curve applies to any ter-molecular reaction in which the initial concentrations are equal. Take (ii.) B in excess, i. e. p= dX (1-X)2(p-X) b α > 1. We have p(1-X)_X(1-p)} 1-X S' log. P(1 − X) Kt= 0 X 0 1 .2 .3 ⚫4 Kt...... 0 028 064 111 ·5 .6 7 .8 ·9 1.0 177 274 416 665 1.07 2.7 ∞0 These figures and the graphs plotted from them (fig. 2) will apply to any ter-molecular reaction of the type 2A + B→1 or more resultants, B being in excess. α Take (iii.) A in excess, and let =p>1, then we have b where K-kb2. log• p(1—X) S' These figures and the corresponding graphs (fig. 3) apply to any ter-molecular reaction of the type 2A+B+1 or more resultants, A being in excess. Quadri-molecular Reaction. Let the reaction be of the type 3A + B→1 or more resultants, and let the initial concentrations be equal. The general X............ 0 1 .2 .3 ·4 .5 .6 Kt 65 .7 1.0 0 123 32 64 1.21 2:33 4.87 7:43 12.0 ∞ b Now take B in excess, putting -=p>1. The equation becomes a X CS 1 = − A log. (1−X) +B: + -- x - 1 } 1-X 2 (1 where A, B, C, D are constants depending on the value of p. p=2. Then A=1. B=-1. 0=1. D=-1. 1.0 These figures give the graphs (fig. 4) for a quadrimolecular reaction of the type 3A+B+, the substance B being in excess. a Now let the substance A be in excess, so that =p>1. A, B, C, D being constants depending on p. p=1. Then A=-64. B=-16. C-4. D=64. 1 (p-X)2 } - Dlog. (1−X), The graphs for these are shown in fig. 5. There is yet the quadri-molecular reaction of the type 2A+2B-1 or more resultants. In this case we have where Kka3 and p= The solution is The general curves for this type of quadri-molecular reaction are shown in fig. 6. Phil. Mag. S. 6. Vol. 35. No. 207. March 1918. X Application of the Curves. (i.) To find the velocity coefficient k. When the order of the reaction and the initial concentrations are known we have only to measure the fraction (X) changed in a given time t in order to find Kt from the curve and therefore k. (ii.) To find the fraction (X) changed in a given time t. This requires a knowledge of k, the order of the reaction, and the initial concentrations. Then X for a given t can be read straight from the curve. (iii.) To find the Order of a reaction. Two or more determinations of X and t are needed together with a knowledge of the initial concentrations. The particular curve on which the points (X, t) best lie, determines the order of the reaction. London, December 1917. XXXII. On the Coefficients of Potential of Two Conducting Spheres. By Prof. A. ANDERSON*. IT T may be of interest to show how the coefficients of potential of two conducting spheres may be obtained directly without a previous determination of the coefficients of capacity and induction, and without making use of electric images. For this purpose the following elementary proposition, which is easily seen to be true, may be used. If a conducting sphere whose radius is a have a charge E, and if other charged bodies be brought into the field, the potential V of the charge E at an external point P whose distance from the centre of the sphere is r will be given by the equation rV=E+a(U−U'), where U is the potential at the centre of the sphere of the introduced charges, and U' their potential at P', the inverse point of P in the sphere. Let A and B be the centres of two spheres whose radii are a and b, the distance AB being c. Let I, be the inverse point of A in the sphere B, I, the inverse of I, in A, I, the inverse of 12 in B, and so on. Also, let J1 be the inverse of B in the sphere A, J, the inverse of J, in B, J; the inverse of J, in A, and so on. 2 1 3 Let U be the potential of the sphere A at B, U1, U2, U3, &c. its potentials at I1, 12, 13, &c., and U'', U,', U', &c. its potentials at J1, J2, J3, &c. Also, let V be the potential of the sphere B at A, V1, V2, V3, &c. its potentials at I1, 12, 13: * Communicated by the Author. |