Use standard heats of formation in Appendix L to calculate standardenthalpy chan
ID: 678417 • Letter: U
Question
Use standard heats of formation in Appendix L to calculate standardenthalpy changes for the following reactions. (a) 0.058 g of sulfur burns,forming SO2(g)kJ
(b) 0.44 mol ofHgO(s) decomposes to Hg(l) andO2(g)
kJ
(c) 2.40 g ofNH3(g) is formed from N2(g)and excess H2(g)
kJ
(d) 1.95 10-2 mol of carbonis oxidized to CO2(g)
kJ
(a) 0.058 g of sulfur burns,forming SO2(g)
kJ
(b) 0.44 mol ofHgO(s) decomposes to Hg(l) andO2(g)
kJ
(c) 2.40 g ofNH3(g) is formed from N2(g)and excess H2(g)
kJ
(d) 1.95 10-2 mol of carbonis oxidized to CO2(g)
kJ
Explanation / Answer
S + O2 ------> SO2 1 mole 1mole 1 mole 0.0018 mole 0.0018 mole .0018 mole mole = weight of thesubstance / Atomic weight Hrxn. = Hprod. - Hreac. = 0.0018 x -296.83 KJ - 0.0018 x 0.33KJ considering form of the sulphur. = -0.538598375 KJ 2 HgO -------> 2 Hg + O2 2moles 2 moles 1 mole 0.44moles 0.44 moles 0.22 moles Hrxn. = Hprod. - Hreac. = 0.44 x 0 + 0.22 x 0 - 0.44 x -90.83 KJ = + 39.9652 KJ N2 + 3 H2 ------> 2NH3 0.070 0.211 0.1411 Hrxn. = Hprod. - Hreac. = 0.1411 x -46.11 KJ - ( 0 ) KJ = - 6.5061 KJ C + O2 --------> CO2 1.95 x 10-2 1.95 x10-2 1.95 x 10-2 Hrxn. = Hprod. - Hreac. = 1.95 x 10-2 x -393.51 KJ = - 7.6734 KJ Hrxn. = Hprod. - Hreac. = 1.95 x 10-2 x -393.51 KJ = - 7.6734 KJRelated Questions
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