Question

Enthalpy of Formation

Part 1:

A scientist measures the standard enthalpy change for the following reaction to be 591.0 kJ : 2H2O(1)—>2H2(g) + O2(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H2O(l) is kJ/mol.

Part 2:

Part 3:

Answer 1

Standard enthalpy of formationen The change of enthalpy to produce a more of a substance at a particular tempt and pressure) from it's constituent pure sta substance (which are also at particular tempe tempt and 1 atm and s atm pr.). part-1 2H₂0 (1) 2 H₂(g) + O2 (9) 4 H = 591.OKJ. > 2H2(g) +0,19) -> 2H2O(l), 40- 5910KJ. H2(g) + 2 O2 (g) - H2011), aki = -501 = - 295.5kJ, A more of H20lk) | So, the standard enthalpy of H₂O (0) - 295.5 KJl more. formation of Part-2 2 C2H6 (9) +702(g) 4 (0,18 + 6H₂0 (9 4HO2-290314KT. So, AH = 44" Hq [log (g)] to go Hg [H20 (9] - 24° Hy [st (8] - 7 g Hj [0₂ (9)] 7 - 2903.4 = 4X(-393.50 +61 H4 [H₂O(g)] - 2x(-88-75) -7xo -- 2903.4 .45 - 1574 + - 6 444 [H₂O(9)] + 167.5.

$60944 [H2009)] = -2.903.4 +1574 -167.5 6 aty [H2O(g)] = -1496.9 at 25°C ► apung [4, 0 (] = -249.48 aty [02(g)]=0 a hf [cog] =-393.5 kJ/more Lain [SHG(8)] = - 83.75 kJ/mer ... Standard enthalpy of formation of H20 (8) KJ/molu. Part-3 2 Nars) + 2H20 (1) 2 Naot lag) + H₂ (8) 44 = -350 kJ. H = 2 4 H4 [Nash (aq) + 4 ty [H₂(g)] ~ 22 ty [ Na(s)] - 24 Hf [H264] = - 350 = 2X (-469.2) +0 -0-2 8 Hf [ H2o(a)] 7) - 24#[W2066)] = -350 + 938.4 A you. - - 2 4 HJ [H20 (0] = +588.4 sty [Na oh (aq)] = -469.2 Ky more >> 3 [H20() = - 29412. 44 [H2 (91] =0 10th [Na(6)]=0. standand enthalp. somation of H20 (1) = -29412 KJ more.
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