Question

problem 5.72 please assist with parts a,b,c and d.

Using values from Appendix C or the internet calculate the value of ΔH∘ for each of the following reactions.

a. 4HBr(g)+O2(g)→2H2O(l)+2Br2(l) express answer using five sig figs ΔH∘rxn =...............kJ

b. 2NaOH(s)+SO3(g)→Na2SO4(s)+H2O(g)2NaOH(s)+SO3(g)→Na2SO4(s)+H2O(g) (ΔHf∘ Na2SO4Na2SO4 = -1387.1 kJ/molkJ/mol)

ΔH∘rxn =...........kJ answer using four sig figs

c. CH4(g)+4Cl2(g)→CCl4(l)+4HCl(g) ΔH∘rxn..............kJ express answer in four sig figs

d. Fe2O3(s)+6HCl(g)→2FeCl3(s)+3H2O(g) ΔH∘rxn..............kJ express answer in two sig figs

Answer 1

Ans.

a) -176.89 KJ/mol

b) -381.98 KJ/mol

c) -429.858 KJ/mol

d) -146.392 KJ/mol

∆H°(rxn) = ∆H°products - ∆H°reactants

NaOH(s) = - 425.609 So3(g)= – 395.72 che values of Horog – 36.4 O2 (g) 249.17 How – 285.83 Bro clag) = 0 HOGE – 92.307 recist – 399.49 Nagsoy (5) – 1387.1 chy g - 24.81 celuces - – 135.44 fego₂ (5) = -824.2. Horg) = -241.818 a SHRxn. Ship-SR SHRxn 2 DH (H2O) + 2 Shf (Bog) - 4 Atif (HiB) - AHF (02) - - 571166 + 394.77 = 176.89 Kijowe b. SHRAM - = WherNag soy) & CHE (H2O) ~2611f (NowH) - Latif (503) -1628,918 +1 246.932 ~381.98 kJ/rwe C. DHRxn & Alif (cale) + 4 or CHCl) - wtf (ory) 4014 (2) - - 504.668 +74.81 – – 429.858 kJ/mol I d. CHRxn - 28 H/C feels) + 3 stig (Ho) - Arij (F303) de – 6 Olif (HCl) - - 1524.434 + 1378.042 = -146.392 kJ/mol