1. The problem statement, all variables and given/known data Calculate the flame temperature of normal octane (liquid) burning in air at an equivalence ration of 0.5. All reactants are at 298 K and the system operates at the pressure of 1 atm. 2. Relevant equations Hp = Hr 3. The attempt at a solution I know that the combustion of liquid octane in the theoretical amount of air is C8H18 + 12.502 + 47N2 --> 8C02 + 9H20 + 47N2 Does the equivalency ratio affect this equation? I know equivalency ratio is the actual fuel/oxidizer ratio divided by the stoichiometric fuel/oxidizer ratio, but what does this mean for the above equation? If we assume the stoichiometric fuel/oxidizer ratio is 1, this means that the actual fuel/oxidizer ratio is 1/2.... If I can get the equation right, I know how to solve for flame temp. using Hp = Hr, JANAF tables, and iteration. Thanks!