Diborane is a potential rocket fuel which undergoes combustion according to the reaction $ {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) + \dfrac{3}{2}{O_2}(g) \to {B_2}{O_3}(s),{\text{ }}\Delta H = - 1273{\text{ }}kJ{\text{ }}mo{l^{ - 1}} $
$ {H_2}(g) + \dfrac{1}{2}{O_2}(g) \to {H_2}O(l),{\text{ }}\Delta H{\text{ = - 286 kJ }}mo{l^{ - 1}} $
$ {H_2}O(l) \to {H_2}O(g),{\text{ }}\Delta H = 44{\text{ }}kJmo{l^{ - 1}} $
$ 2B(s) + 3{H_2}(g) \to {B_2}{H_6}(g),{\text{ }}\Delta H = 36{\text{ }}kJ{\text{ }}mo{l^{ - 1}} $
Last updated date: 24th Mar 2023
•
Total views: 207k
•
Views today: 1.84k
Answer
207k+ views
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.
$ \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:
$ {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:
$ \Delta H = + (\Sigma (\Delta {H_f}{\text{ of products) - }}\Sigma {\text{(}}\Delta {{\text{H}}_f}{\text{ of reactants))}} $
$ = + [(\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))] $
$ {H_2}(g) + \dfrac{1}{2}{O_2}(g) \to {H_2}O(l),{\text{ }}\Delta H{\text{ = - 286 kJ }}mo{l^{ - 1}} $
$ {H_2}O(l) \to {H_2}O(g),{\text{ }}\Delta H = 44{\text{ }}kJmo{l^{ - 1}} $
$ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ $
$ {H_2}(g) + \left( {\dfrac{1}{2}} \right){O_2}(g) \to {H_2}O(g) $
$ \Delta {H_f} = - 242kJmo{l^{ - 1}} $
After substituting the given data,
$ \Delta H = [( - 1273) + (3 \times - 242)] - (36 - 3 \times 0) + [ - 1999 - 36] $
$ = - 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.
$ \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:
$ {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:
$ \Delta H = + (\Sigma (\Delta {H_f}{\text{ of products) - }}\Sigma {\text{(}}\Delta {{\text{H}}_f}{\text{ of reactants))}} $
$ = + [(\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))] $
$ {H_2}(g) + \dfrac{1}{2}{O_2}(g) \to {H_2}O(l),{\text{ }}\Delta H{\text{ = - 286 kJ }}mo{l^{ - 1}} $
$ {H_2}O(l) \to {H_2}O(g),{\text{ }}\Delta H = 44{\text{ }}kJmo{l^{ - 1}} $
$ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ $
$ {H_2}(g) + \left( {\dfrac{1}{2}} \right){O_2}(g) \to {H_2}O(g) $
$ \Delta {H_f} = - 242kJmo{l^{ - 1}} $
After substituting the given data,
$ \Delta H = [( - 1273) + (3 \times - 242)] - (36 - 3 \times 0) + [ - 1999 - 36] $
$ = - 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.
Recently Updated Pages
In India on the occasion of marriages the fireworks class 12 chemistry JEE_Main

The alkaline earth metals Ba Sr Ca and Mg may be arranged class 12 chemistry JEE_Main

Which of the following has the highest electrode potential class 12 chemistry JEE_Main

Which of the following is a true peroxide A rmSrmOrm2 class 12 chemistry JEE_Main

Which element possesses the biggest atomic radii A class 11 chemistry JEE_Main

Phosphine is obtained from the following ore A Calcium class 12 chemistry JEE_Main
