Use standard heats of formation in Appendix L to calculate standardenthalpy chan

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 KJ        

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