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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)):}`

A

203KJ

B

216KJ

C

330KJ

D

343KJ

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The correct Answer is:
a
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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):}

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

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)

Compounds with carbon-carbon double bond, such as ethylene, C_(2)H_(4) , add hydrogen in a reaction called hydrogenation. C_(2)H_(4)(g)+H_(2)(g) rarr C_(2)H_(6)(g) Calculate enthalpy change for the reaction, using the following combustion data C_(2)H_(4)(g) + 3O_(2)(g) rarr 2CO_(2)(g) + 2H_(2)O(g) , Delta_("comb")H^(Θ) = -1401 kJ mol^(-1) C_(2)H_(6)(g) + 7//2O_(2)(g)rarr 2CO_(2) (g) + 3H_(2)O(l) , Delta_("comb")H^(Θ) = -1550kJ H_(2)(g) + 1//2O_(2)(g) rarr H_(2)O(l) , Delta_("comb")H^(Θ) = -286.0 kJ mol^(-1)

Calculate the standard enthalpy of formation of acetylene from the following data : C_((g))+O_(2(g))rarr CO_(2(g)),DeltaH^(@)=-393kJ mol^(-1) H_(2(g))+(1)/(2)O_(2(g)) rarr H_(2)O_((l)),DeltaH^(@)=-285.8kJ mol^(-1) 2C_(2)H_(2(g))+5O_(2(g))rarr 4CO_(2(g))+2H_(2)O_((l)),DeltaH^(@)=-2598.8kJ mol^(-1) .

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