Question

use the table below to calculate ΔHo reaction:


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

Answer 1

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.