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Question: Diborane is a potential rocket fuel which undergoes combustion according to the reaction \( {B_2}{H_...

Diborane is a potential rocket fuel which undergoes combustion according to the reaction B2H6(g)+3O2(g)→B2O3(s)+3H2O(g){B_2}{H_6}(g) + 3{O_2}(g) \to {B_2}{O_3}(s) + 3{H_2}O(g)
From the following data, calculate the enthalpy change for the combustion of diborane.
2B(s)+32O2(g)→B2O3(s), ΔH=−1273 kJ mol−12B(s) + \dfrac{3}{2}{O_2}(g) \to {B_2}{O_3}(s),{\text{ }}\Delta H = - 1273{\text{ }}kJ{\text{ }}mo{l^{ - 1}}
H2(g)+12O2(g)→H2O(l), ΔH = - 286 kJ mol−1{H_2}(g) + \dfrac{1}{2}{O_2}(g) \to {H_2}O(l),{\text{ }}\Delta H{\text{ = - 286 kJ }}mo{l^{ - 1}}
H2O(l)→H2O(g), ΔH=44 kJmol−1{H_2}O(l) \to {H_2}O(g),{\text{ }}\Delta H = 44{\text{ }}kJmo{l^{ - 1}}
2B(s)+3H2(g)→B2H6(g), ΔH=36 kJ mol−12B(s) + 3{H_2}(g) \to {B_2}{H_6}(g),{\text{ }}\Delta H = 36{\text{ }}kJ{\text{ }}mo{l^{ - 1}}

Explanation

Solution

Hint : Enthalpy change is the heat passing into or out of the system during a reaction. The heat content of a system is the enthalpy. The enthalpy change of a chemical reaction is roughly equal to the amount of energy lost or gained during the chemical reaction.
ΔH=+(Σ(ΔHf of products) - Σ(ΔHf of reactants))\Delta H = + (\Sigma (\Delta {H_f}{\text{ of products) - }}\Sigma {\text{(}}\Delta {{\text{H}}_f}{\text{ of reactants))}}

Complete Step By Step Answer:
Diborane is a potential fuel and it undergoes combustion reaction as below:
B2H6(g)+3O2(g)→B2O3(s)+3H2O(g){B_2}{H_6}(g) + 3{O_2}(g) \to {B_2}{O_3}(s) + 3{H_2}O(g)
To calculate the enthalpy change for the combustion reaction of diborane, use the following formula:
ΔH=+(Σ(ΔHf of products) - Σ(ΔHf of reactants))\Delta H = + (\Sigma (\Delta {H_f}{\text{ of products) - }}\Sigma {\text{(}}\Delta {{\text{H}}_f}{\text{ of reactants))}}
=+[(ΔHf of B2O3(g)+3×ΔHf of H2O(g)) - (ΔHf of B2H6(g)+3×ΔHf of O2(g))]= + [(\Delta {H_f}{\text{ of }}{{\text{B}}_2}{{\text{O}}_3}(g) + 3 \times \Delta {H_f}{\text{ of }}{{\text{H}}_2}{\text{O(g)) - (}}\Delta {{\text{H}}_f}{\text{ of }}{{\text{B}}_2}{{\text{H}}_6}(g) + 3 \times \Delta {H_{f{\text{ }}}}{\text{of }}{{\text{O}}_2}(g))]
H2(g)+12O2(g)→H2O(l), ΔH = - 286 kJ mol−1{H_2}(g) + \dfrac{1}{2}{O_2}(g) \to {H_2}O(l),{\text{ }}\Delta H{\text{ = - 286 kJ }}mo{l^{ - 1}}
H2O(l)→H2O(g), ΔH=44 kJmol−1{H_2}O(l) \to {H_2}O(g),{\text{ }}\Delta H = 44{\text{ }}kJmo{l^{ - 1}}
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H2(g)+(12)O2(g)→H2O(g){H_2}(g) + \left( {\dfrac{1}{2}} \right){O_2}(g) \to {H_2}O(g)
ΔHf=−242kJmol−1\Delta {H_f} = - 242kJmo{l^{ - 1}}
After substituting the given data,
ΔH=[(−1273)+(3×−242)]−(36−3×0)+[−1999−36]\Delta H = [( - 1273) + (3 \times - 242)] - (36 - 3 \times 0) + [ - 1999 - 36]
=−2035kJmol−1= - 2035kJmo{l^{ - 1}}

Note :
The enthalpy change which occurs when one mole of a compound is burnt completely in oxygen under standard conditions and with everything in standard state is called the standard enthalpy change of combustion of the compound. Heat is released in combustion reactions. So, the total heat content should decrease in combustion reactions.