The fat, `C_(57)H_(104)O_(6)(s)`, is metabolized via the following reaction. Given the enthalpies of formation, calculate the energy (kJ) liberated when 1.0 g of this fat reacts. `C_(57)H_(104)O_(6)(s)+80 O_(2)(g)rarr57CO_(2)(g)=-52 H_(2)O(l)` `Delta_(f)H^(@)(C_(57)H_(104)O_(6),s)=-70870" kJ"//"mol, "Delta_(f)H^(@)(H_(2)O,l)=-285.8 " kJ"//"mol"`, `Delta_(f)H^(@)(CO_(2),g)=-393.5 " kJ"//"mol"`
A
37.98
B
40.4
C
33.4
D
30.2
Text Solution
Verified by Experts
The correct Answer is:
A
`Delta_(r )H^(@)=57 xx (-393.5)+52xx(-285.8)-(-70870)` `=-33578.9" kJ"//"mol"` `=-(33578.9)/(884)=-37.98 " kJ"//"mol"`
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Determine the enthalpy of formation of B_(2)H_(6) (g) in kJ/mol of the following reaction : B_(2)H_(6)(g)+3O_(2)(g)rarrB_(2)O_(3)(s)+3H_(2)O(g) , Given : Delta_(r )H^(@)=-1941 " kJ"//"mol", " "DeltaH_(f)^(@)(B_(2)O_(3),s)=-1273" kJ"//"mol," DeltaH_(f)^(@)(H_(2)O,g)=-241.8 " kJ"//"mol"
Calculate Delta_(r) H^(@) for Fe_(2) O_(3) (s) + 3 CO(g) to 2 Fe (s) + 3 CO_(2) (g) Given : Delta_(f) H^(@) (Fe_(2) O_(3) , s) = -822.2 kJ/mol Delta_(f) H^(@) (CO , g) = -110.5 kJ//mol , Delta_(f) H^(@) (CO_(2) , g) = -393.5 kJ/mol
Calculate the Delta_(r)G^(@) for the reaction: C_(6)H_(12)O_(6)(s) + 6O_(2)(g) to 6CO_(2)(g) + 6H_(2)O(l) Delta_(f)G^(@) values (kJ mol^(-1) ) are : C_(12)H_(12)O_(6)(s) = -910.2, CO_(2)(g) = -394.4, H_(2)O(l) = -237.2
Determine C-C and C-H bond enthalpy (in kJ//mol ) Given : Delta_(f)H^(@)(C_(2)H_(6),g)=-85kJ//mol," "Delta_(f)H^(@)(C_(3)H_(8),g)=-104kJ//mol Delta_("sub")H^(@)(C,s)=718kJ//mol," "B.E.(H-H)=436kJ//mol
Calculate standard free energy change from the free energy of formation data for the following reaction, C_(6)H_(6)(l)+(15)/(2)O_(2)(s)to 6CO_(2)(g)+3H_(2)O(g) . Delta_(f)G^(theta)[C_(6)H_(6)(l)]=172.8 kJ/mol Delta_(f)G^(theta)[CO_(2)(g)]=-394.4 kJ/mol Given that. Delta_(f)G^(theta)[CO_(2)(g)]=394.4 kJ/mol Delta_(r )G^(theta)[H_(2)O(l)]=-228.6 kJ/mol
Calculate the standard Gibbs free energy change from the free energies of formation data for the following reaction: C_(6)H_(6)(l) +(15)/(2)O_(2)(g) rarr 6CO_(2)(g) +3H_(2)O(g) Given that Delta_(f)G^(Theta) =[C_(6)H_(6)(l)] = 172.8 kJ mol^(-1) Delta_(f)G^(Theta)[CO_(2)(g)] =- 394.4 kJ mol^(-1) Delta_(f)G^(Theta) [H_(2)O(g)] =- 228.6 kJ mol^(-1)
Calculate the DeltaH_(f)^@ of C_(6)H_(12)O_(6) (s) from the following data: DeltaH_(comb)[C_(6)H_(12)O_(6)(s)]=-2816.kJ//mol DeltaH_(f)^(@) of CO_(2)(g)=-393.5kJ//mol DeltaH_(f)^(@) of H_(2)O(l)=-285.9kJ//mol
Calculate the enthalpy of formation of ethyl alcohol from the following data : C_(2)H_(5)OH(l) + 3O_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l), Delta_(r)H^(@) = -1368.0 kJ C(s) +O_(2)(g) to CO_(2)(g), Delta_(r)H^(@) = -393.5 kJ H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l), Delta_(r)H^(@) = -286.0 kJ
Use the information in the table to calculate the enthalpy of this reaction . C_(2)H_(6)(g)+(7)/(2)O_(2)(g)to2CO_(2)(g)+3H_(2)O(l) {:(Reaction ,DeltaH_(f)^(@)KJ.mol^(-1)),(2C(s)+3H_(2)(g)toC_(2)H_(6)(g),-84.7),(2C(s)+O_(2)(g)toCO_(2)(g),-393.5),(H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),-285.8):}
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