standard enthalpy change of formation

Question

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Using the values of standard enthalpy change of formation given calculate the values of△H for the following reactions:

a. the thermal decomposition of BaCO3
b. C2H6(g) +Br2(l)  C2H5Br(l)+ HBr(g)
c. C2H4(g) + H2O(l) C2H5OH(l)
d. TiCl4(l) +2Mg(s)  Ti(s)+2MgCl2(s)
e. SO3(l) + H2O(l) H2SO4(l)

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[△Hf/kJmol-1: BaO(s): -997.5; BaCO3(s): -1218.8; CO2(g): -393.5; C2H4(g): +52.3; C2H6(g): -84.6; C2H5Br(l): -92.0; C2H5OH(l): -277.7; HBr(g): -36.2; H2O(l): -285.9; H2SO4(l): -814.0; MgCl2(s): -641.8; TiCl4(l): -804.2; SO3(l): -395.4]

Answer 1

For any reaction,
ΔH = ∑ΔHf(products) - ∑ΔHf(reactants)

=====
a.
BaCO3(s) → BaO(s) + CO2(g)
ΔH
= {ΔHf[BaO(s)] +ΔHf[CO2(g)]} - ΔHf[BaCO3(s)]
= {(-997.5) + (-393.5)} - (-1218.8)
= -172.2 kJ mol-1

=====
b. ΔH
= {ΔHf[C2H5Br(l)] +ΔHf[HBr(g)]} - {ΔHf[C2H6(g)] +ΔHf[HBr(g)]}
= {(-92.0) + (-36.2)} - {(-84.6) + 0}
= -43.6 kJ mol-1

=====
c. ΔH
= ΔHf[C2H5OH(l)] - {ΔHf[C2H4(g)] +ΔHf[H2O(l)]}
= (-277.7) - {(+52.3) + (-285.9)}
= -44.1 kJ mol-1

=====
d. ΔH
= {ΔHf[Ti(s)] +2ΔHf[MgCl2(s)]} - {ΔHf[TiCl4(l)] +2ΔHf[Mg(s)]}
= {0 + 2(-641.8)} - {(-804.2) + 2(0)}
= -479.4 kJ mol-1

=====
e. ΔH
= ΔHf[H2SO4(l)] - {ΔHf[SO3(l)] +ΔHf[H2O(l)]}
= (-814.0) - {(-395.4) + (-285.9)}
= -132.7 kJ mol-1