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Calcualte the amount of energy released ...

Calcualte the amount of energy released when 0.100 mol of diborane ,`B_(2)H_(6)` , reacts with oxygen to produce soild `B_(2)O_(3)`and steam:
`{:overset(Delta_(f)H(kJ.mol^(-1)))((B_(2)H_(6)(g),35),(B_(2)O_(3)(g),-1272),(BH_(2)O(l),-285),(H_(2)O(g),-241)):}`

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Diborane (B_(2)H_(6)) reacts independently with O_(2) and H_(2)O to produce, respectively:

Calculate the bond energy of O-H bond in H_(2)O(g) at the standard state from the following data: (1) H_(2)(g) to 2H(g),DeltaH^(0)=436kJ*mol^(-1) (2) (1)/(2)O_(2)(g)toO(g),DeltaH^(0)=249k*mol^(-1) (3) H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(g),DeltaH_(f)^(0)[H_(2)O(g)]=-241.8kJ*mol^(-1) .

Diborane is a potential rocket fuel which undergoes combustion according to the reaction, B_(2)H_(6) (g) + 3O_(2) (g) rarr B_(2) O_(3) (s) + 3H_(2) O (g) From the following data, calculate the enthalpy change for the combustion of diborane: (i) 2 B (s) + ((3)/(2)) O_(2) (g) rarr B_(2) O_(3) (s) , Delta H = 0 1273 k J//mol (ii) H_(2) (g) + ((1)/(2)) O_(2) (g) rarr H_(2) O (l), Delta H = -286 kJ//mol (iii) H_(2) O (l) rarr H_(2)O (g) Delta H = 44 kJ//mol (iv) 2B (s) + 3H_(2) (G) rarr B_(2) H_(6) (g), Delta H = 36 kJ//mol

Calculate the enthalpy of combustion of ethylene . Given : C_(2) H_(6) (g) + (7)/(2) O_(2) (g) to 2 CO_(2) + 3 H_(2)O (l) , Delta H = = 1562 kJ* mol^(-1) H_(2) (g) + (1)/(2) O_(2) (g) to H_(2) O (l) , Delta H = -286 kJ * mol^(-1) C_(2) H_(4) (g) + H_(2) (g) to C_(2) H_(6) (g) , Delta H = -32 kJ * mol^(-1)

Diborane is a potentail rocket fuel which undergoes combusion according to the reaction B_(2)H_(6)(g)+3O_(2)(g)rarrB_(2)O_(3)(s)+3H_(2)O(g) From the following data, calculate the enthalpy change for the combustion of diborane 2B(s)+(3//2)O_(2)(g)rarrB_(2)O_(3)(s) DeltaH= -1273 kJ mol^(-1) H_(2)(g)+(1//2)O_(2)(g)rarrH_(2)O(1) " "Delta H= -286 kJ mol^(-1) H_(2)O(1)rarrH_(2)O(g)" "DeltaH=44 kJ mol^(-1) 2B(s)+3H_(2)(g)rarrB_(2)H_(6)(g)" "Delta H= 36kJ mol^(-1)

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