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Find the enthalpy of S-S bond from the f...

Find the enthalpy of `S-S` bond from the following data.
`{:((i),C_(2)H_(5)-S-C_(2)H_(5) (g),Delta_(f)H^(@)=-147.2" kJ/mol"),((ii),C_(2)H_(5)-S-S-C_(2)H_(5)(g),Delta_(f)H^(@)=-201.9" kJ/mol"),((iii),S(g),Delta_(f)H^(@)=222.8" kJ/mol"):}`

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The correct Answer is:
277.5 kJ/mol
`C_(2)H_(5) - S - C_(2) H_(5) + S rarr C_(2)H_(5) - S - S - C_(2) H_(5)`
`Delta H = B.E.` of `S - S` bond
`= - 201.9 - [222.8 - 147.2] = - 277.5`
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The bond dissociation energy of a diatomic molecule is also called bond energy. However, the bond dissociation energy depends upon the nature of bond and also the molecule in which the bond is present. The bond energy N-H bind in NH_(3) is equal to one-third of the energy of dissociation of NH_(3) because there are three N-H bonds and those of C-H bond in CH_(4) is equal to one-fourth of the energy of dissociation of CH_(4) Heat of a reaction = Bond energy of reactants - Bond energy of products Find the bond enthalpy of S-S bond from the following data. C_(2)H_(5)-S-C_(2)H_(5)(g) " " Delta H_(f)^(@) = -147.23 kJ mol^(-1) C_(2)H_(5)-S-S-C_(2)H_(5)(g) " " Delta H_(f)^(@) = -201.92 kJ mol^(-1) S(g) Delta H_(f)^(@) = 222.80 kJmol^(-1)

Find bond enthalpy of S-S bond from the following data: C_(2)H_(5)-S-C_(2)H_(5)," "DeltaH_(f)^(@)=-147.2kJ" "mol^(-1) C_(2)H_(5)-S-S-C_(2)H_(5)," "DeltaH_(f)^(@)=-201.9kJ" "mol^(-1) S(g)," "DeltaH_(f)^(@)=222.8kJ" "mol^(-1)

Find the enthalpy of S-S bond from the following data. Report your answer by multiplying 1.08/100 (i) C_2H_5-S-C_2H_5(g) " " DeltaH_f^(2)=-147.2kJ//mol (ii) C_2H_5-S-S-C_2H_5(g) " " DeltaH_f^(@)=-201.9 kJ//mol (iii) S(g) " "DeltaH_f^(@)=222.8kJ//mol

Calculate the enthalpy of formation of ethane from the following data : (i) C(s) + O_(2)(g) to CO_(2)(g) , Delta_(r)H^(@) = -393.5 kJ (ii) H_(2)(g) + 1/2 O_(2)(g) to H_(2)O(l), Delta_(r)H^(@) = -285.8 kJ (iii) C_(2)H_(6)(g) + 7/2O_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l), Delta_(r)H^(@) = -1560.0 kJ

Calculate the lattice enthalpy of MgBr_(2) from the given date: {:(Mg(s)+Br_(2)(l)rarrMgBr_(2)(s),,Delta_(f)H^(@)=-524 kJ mol^(-1)),(Mg(s)rarrMg(s),,Delta_(1)H^(@)=+148 kJ mol^(-1)),(Mg(g)rarrMg^(2+)(g)+2e^(-),,Delta_(2)H^(@)=+2187 kJ mol^(-1)),(Br_(2)(l)rarrBr_(2)(g),,Delta_(2)H^(@)=+2187 kJ mol^(-1)),(Br_(2)(l)rarrBr_(2)(g),,Delta_(3)H^(@)=+31 kJ mol^(-1)),(2Br(g)+2e^(-)rarr2Br(g),,Delta_(5)H^(@)=-662 kJ mol^(-1)):} Strategy : The thermochemical equation corresponding to lattice enthalpy of MgBr_(2) is {:(Mg^(2+)(g)+2Br^(-)(g)rarrMgr_(2)(s),,,Delta_("Lattice")H^(@)=?):} Add the last five thermochemical equations to the thermochemical equation corresponding to lattice enthalpy to get the thermochemical equation for the formation of MgBr_(2) (s) from its constituent element. Finally, calculate Delta_(Lattice) H^(@) , using the concept of Hess's law.

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