Please check my logic and math...

I need to calculate the heat of reaction for citric acid and sodium carbonate. Like Mr. Jones, I couldn't find the heat of formation for sodium citrate. I did, however, find the heat of neutralization for citric acid with NaOH. (see:

http://www.ecama.org/level_2/frames/safety_fr.html, click on citric acid; the math: 265 cal/gm = 0.265 * 192 = 50.9 kcal/gm-mole).

Knowing the heat of formation of NaOH (-112.2 kcal/gm-mole), water (-68.3 kcal/gm-mole) and citric acid (-369 kcal/gm-mole), I back calculated the heat of formation of trisodium citrate as -551.8 cal/gm-mole:

citric acid + 3NaOH ---> Na3 citrate + 3H2O

50.9=(x+(3*-68.3))-(-369+(3*-112.2))

x=-551.8

I plugged this value into the stoichiometry of the citric acid/Na2CO3 reaction (knowing the heat of formation for CO2 is -94 kcal/gm-mole and Na2CO3 is -275 kcal/gm-mole):

2 Citric Acid + 3 Na2CO3 ---> 2Na3citrate + 3 CO2 + 3 H2O

({2*-551.8} + {3*-94} + {3*-68.3}) - ({2*-369} + {3*-275})=27.1 (endothermic)

Is my math and 30+ year old chemistry right??? If not, please correct!

Thanks!

Arau