Use standard heats of formation listed below to calculatestandard enthalpy chang

Question

Use standard heats of formation listed below to calculatestandard enthalpy changes for the following: sulfur =0kJ/mol                   SO2=-296.84 kJ/mol HgO= -90.83kJ/mol            Hg=0kJ/mol              O2=0kJ/mol NH3=-45.90kJ/mol              N2= 0kJ/mol            H2=0kJ/mol          C=0kJ/mol                          CO2=-393.509 kJ/mol a)0.054 g of sulfur burns, forms SO2(g) b) 0. 20 mol of HgO(s) decomposes to Hg(l) andO2(g) c) 2.40 g of NH3 is formed from N2(g)and excess H2(g) d) 1.05 x 10-2 mol of carbon is oxidized toCO2(g) Use standard heats of formation listed below to calculatestandard enthalpy changes for the following: sulfur =0kJ/mol                   SO2=-296.84 kJ/mol HgO= -90.83kJ/mol            Hg=0kJ/mol              O2=0kJ/mol NH3=-45.90kJ/mol              N2= 0kJ/mol            H2=0kJ/mol          C=0kJ/mol                          CO2=-393.509 kJ/mol a)0.054 g of sulfur burns, forms SO2(g) b) 0. 20 mol of HgO(s) decomposes to Hg(l) andO2(g) c) 2.40 g of NH3 is formed from N2(g)and excess H2(g) d) 1.05 x 10-2 mol of carbon is oxidized toCO2(g)

Explanation / Answer

We Know that :      S + O2 -------> SO2   a.   The amount of sulphur taken = 0.054g         number of molesof sulphur = 0.054 g / 32 g /mol                                                      =  0.00168 mol         The amount of energy released = -296.84 kJ for 1 mole                Theamount of energy released for 0.00168 mol x-296.84 kJ                                                                 =  0.500 kJ    b.          2 HgO<------> 2 Hg + O2              The amount ofenergy required for the decomposition of 1 molof HgO is -90.83 kJ          The amount ofenergy required for 0.20 mol HgO is 0.20 mol x - 90.83kJ / mol                                                                                        = -18.166 kJ     c.           N2 + 3 H2 <--------> 2 NH3        c.   The weight of NH3 taken = 2.40 g            number of moles of NH3 formed = 2.40 g / 17 g / mol                                                                  = 0.1411 mol             The amount of energy released for the formation of   1mol NH3 is -45.90 kJ               The amount of energy released for the formation of 0.1411 molNH3 is -6.48 kJ       d.                    C + O2 ------> CO2               1.05 x 10-2mol                      The number ofmoles of   CO2 formed is 1.05 x10-2 mol           Theenergy released for the formation of 1 mol of CO2 is -393.509 kJ           The energy released for the formation of 1.05 x10-2 mol of CO2 is -4.131 kJ

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