use the table below to calculate ΔHo reaction:

Question

Standard Enthalpies of Formation, ΔH˚fٍ at 298 K 
Substance - Formula - ΔH˚ƒ(kj/mol) 
Acetylene    C2H2(g)      226.7 
Ammonia      NH3(g)      -46.19 
Benzene      C6H6(l)      49.04 
Calcium Carbonate  CaCO2(s) -1027.1 
Calcium Oxide  CaO(s)     -635.5 
Carbon Dioxide  CO2(g)    -393.5 
Carbon Monoxide  CO(g)     -110.5 
Diamond         C(s)        1.88 
Ethane         C2H6(g)     -84.68 
Ethanol        C2H6OH(l)    -277.7 
Ethylene         C2H4(g)    52.30 
Glucose         C6H12O6      -1260 
Hydrogen Bromide  HBr(g)    -36.23 
Hydrogen Chloride  HCl(g)    -92.30 
Hydrogen Fluoride  HF(g)     -268.6 
Hydrogen Iodide     Hl(g)      25.9 
Methane            CH4(g)    -74.85 
Methanol          CH3OH(l)   -238.6 
Propane            C3H8(g)   -103.85 
Silver Chloride     AgCl(s)   -127.0 
Sodium Bicarbonate  NaHCO3(s) -947.7 
Sodium Carbonate   Na2CO3(s)  -1130.9 
Sodium Chloride      NaCl(s)   -411.0 
Sucrose           C12H22O11(s)   -2221 
Water               H2O(l)     -285.8 
Water Vapor         H2O(g)     -241.8

Determine whether the following are endothermic or exothermic.

2H2O2 (l) → 2H2O(l) + O2(g)
∆H = -190 kj

C(s) + H2O (g) → CO (g) + H2 (g)
∆H = +113 kj

C4H9OH(l) + 6 O2(g) → 4 CO2(g) + 5 H 2O(g)
∆H = -2456.1

2 C(s) + H2(g) → C2H2(g)
∆H = +226.71

N2O5(s) + H2O(l) → 2 HNO3(aq)
∆H = -86.0

DrBob222 · Answer

I don't understand the first part of the question. In fact, I don't even know if it is a question, especially with such a long table for reference. For the second part, if delta H is negative the reaction is exothermic. If delta H is positive the reaction is endothermic.