FLAMES OF ATOMIC HYDROGEN 135 



molecule of gas from room temperature (300° K.) to T at constant 

 pressure is ^^ 



For H (atomic) Qx = 4.97 T — 1490 



H2 Q2 = 6.50 T + 0.00045 T^ ~ 1990 



CO or O- 03 = 6.50 7 -j- 0.00050 T^ — 2000 



H2O Q4 = 8.81 T - 0.00095 T2 + 74 X 10-^ T^ - 2570 



CO2 Q5 = 7-0 T + 0.00355 72 - 6.2 X 10-^ T^ - 2400 



Thus the temperature that could be reached by the recombination of 

 hydrogen atoms starting from room temperature and atmospheric pressure 

 can be calculated by the equation 



(i-x)Ho = 2xQi+(i-x) Q2 (8) 



where Ho is the heat of combination of 2.016 grams of atomic hydrogen 

 at room temperature (98,000 calories). Introducing the values of Qi and 

 Q2 in this equation and solving for x we find 



_ 100,000 — 6.50 r — 0.00045 T^ . . 



97,000 -|- 3.44 T — 0.00045 7^ ^ 



We can thus plot a curve giving x as a function of T. But the data of 

 Table I enable us to plot another curve of ;tr as a function of T. These two 

 curves intersect at T = 4030° K. and x = 0.642. Thus atomic hydrogen 

 at room temperature and atmospheric pressure would heat itself to 4030° K. 

 and the degree of dissociation would then be 0.642. 



The temperature of the oxyhydrogen flame can be calculated in an 

 analogous manner, taking into account the dissociation of the water vapor 

 into oxygen and hydrogen and the dissociation of the molecular hydrogen 

 into atomic. This rather laborious calculation shows that the combustion 

 of a mixture of two volumes of hydrogen with one of oxygen at room 

 temperature (300° K.) and atmospheric pressure produces enough heat 

 to raise the reaction products to 3130° K. The reaction is 



2H2 + O2 = 1. 652 H2O + 0.249 H2 + 0. 198 H + 0.17402 (10) 



Thus the composition (by volume) of the flame gases at this maximum 

 temperature is water vapor 72.68, molecular hydrogen 10.95, atomic 

 hydrogen 8.71, and molecular oxygen 7.66 per cent. At this temperature 

 the equilibrium constant for the dissociation of hydrogen according to the 

 data of Table 1 is K = 0.068 atmosphere. The corresponding equilibrium 

 constant for the dissociation of water vapor at 3130° K. is ^® 



K = (H2)2 (02)/(H20)2 = 0.00170 (11) 



*° The specific heats for these calculations were taken from a table on p. 80 of 

 Lewis and Randall's "Thermodynamics." 



*®Langmuir, Jour. Amer. Chem. Soc, 28, 1357 (1906). 



