The influence of nitric oxide on the two opposing reactions of the equilibrium C 2 H 6 ⇌ C 2 H 4 + H 2

Abstract

The combination of ethylene and hydrogen, like the decomposition of ethane, is reduced to a constant limiting rate, well removed from zero, by the addition of relatively small amounts of nitric oxide. The residual reaction is regarded as a non-chain molecular combination. The following thermodynamic requirements of the system C 2 H 6 ⇌ C 2 H 4 + H 2 are verified (as closely as can be expected in view of an appreciable irreversible formation of methane): ( а ) In the presence of enough nitric oxide the combination rate is proportional, above about 200 mm total pressure, to p c 2 H 4 p H 2 , the ethane decomposition rate being propor­tional to p c 2 H 6 , at low pressures the combination shows an increasing dependence on total pressure, which corresponds to an increase in the order of the reverse reaction. ( b ) In the absence of nitric oxide the chain propagation in the ethane decomposition becomes dependent on p H 3 and p c 2 H 4 , Correspondingly, the order of the combination reaction also changes in the way required to preserve the correct form of the equilibrium constant. ( c ) The forms of the ‘relative inhibition curves’ (fraction of chain reaction left unsup­pressed plotted against nitric oxide) are similar for the two opposed reactions. This is accounted for kinetically if the chain part of the combination takes place essentially by the steps H + C 2 H 4 → C 2 H 5 , C 2 H 5 + H 2 → C 2 H 6 + H, and the chain part of the ethane decomposition by the reverse of these. ( d ) For the two inhibited (molecular) reactions the difference in the activation energies agrees satisfactorily with the heat of reaction. For the total reaction, chain plus molecular, if the activation energies are determined from the initial rates of reaction of the pure gases without allowance for the complex mutual influences which the various gases would have on chain propagation at equilibrium, no simple relation is found.