[friedpork]

Calculate the enthalpy change at 20 degrees C for the reaction
2NaCl(s) + H2SO4(l) = Na2SO4(s) + 2HCl(g)
from the following data obtained at 20 degrees C and 1 atm pressure:
Na(s) + (1/ 2)Cl2(g) = NaCl(s)   H = -411.7 kJ
H2(g) + S(s) + 2O2 (g) = H2SO4(l)   H = -793.7 kJ
2Na(s) + S(s) + 2O2 (g) = Na2SO4(s)   H = -1365.2 kJ
(1/ 2)H2(g) + (1/ 2)Cl2 (g) = HCl(g)   H = -92.0 kJ

I know that H = H(products) - H(reactants)
= H (Na2SO4) + 2 H (HCl) - 2 H (NaCl) - H (H2SO4) according the the answer key

And then I just plug in the numbers and get my answer. But I don't understand why there is a negative sign in front of H (H2SO4). If somebody can explain to me why the sign changes that would be great. THank you

[Big-Daddy]

Calculate the enthalpy change at 20 degrees C for the reaction
2NaCl(s) + H2SO4(l) = Na2SO4(s) + 2HCl(g)
from the following data obtained at 20 degrees C and 1 atm pressure:
Na(s) + (1/ 2)Cl2(g) = NaCl(s)   H = -411.7 kJ
H2(g) + S(s) + 2O2 (g) = H2SO4(l)   H = -793.7 kJ
2Na(s) + S(s) + 2O2 (g) = Na2SO4(s)   H = -1365.2 kJ
(1/ 2)H2(g) + (1/ 2)Cl2 (g) = HCl(g)   H = -92.0 kJ

I know that H = H(products) - H(reactants)
= H (Na2SO4) + 2 H (HCl) - 2 H (NaCl) - H (H2SO4) according the the answer key

And then I just plug in the numbers and get my answer. But I don't understand why there is a negative sign in front of H (H2SO4). If somebody can explain to me why the sign changes that would be great. THank you

H2SO4 is a reactant (shown by your original reaction equation for which you want the enthalpy change) thus it is subtracted.