Question

Part A

Calculate the enthalpy of the reaction

2NO(g)+O2(g)→2NO2(g)

given the following reactions and enthalpies of formation:

12N2(g)+O2(g)→NO2(g),   ΔH∘A=33.2 kJ

12N2(g)+12O2(g)→NO(g),  ΔH∘B=90.2 kJ

Express your answer with the appropriate units.

Part B

Calculate the enthalpy of the reaction

4B(s)+3O2(g)→2B2O3(s)

given the following pertinent information:

B2O3(s)+3H2O(g)→3O2(g)+B2H6(g),    ΔH∘A=+2035 kJ

2B(s)+3H2(g)→B2H6(g),                            ΔH∘B=+36 kJ

H2(g)+12O2(g)→H2O(l),                ΔH∘C=−285 kJ

H2O(l)→H2O(g),                                          ΔH∘D=+44 kJ

Express your answer with the appropriate units.

Answer 1

Concepts and reason

An equation which contains all the reactants and products with their physical states mentioned in brackets and enthalpy of reaction is termed as thermochemical equation.

The enthalpy of a reaction is the change in energy during the reaction at given temperature and pressure.

The enthalpy of given reaction is calculated using Hess’s law by taking help of the reaction enthalpies of other given reactions.

Fundamentals

Hess’s law states that the total enthalpy change for a reaction during the complete course of time is same irrespective of the path it follows.

If a reaction is made up of more than one step, the overall enthalpy of the final reaction is the summation of individual enthalpies of each step.

The thermochemical equation follows some simple set of rules to calculated the enthalpy of a reaction:

1. If a reaction is reversed, the enthalpy change sign is also reversed.

2. If a reaction is multiplied/divided by a constant, the enthalpy is also multiplied/divided with that same number.

Part A

The given reactions are as follows:

$\frac{1}{2}{{\rm{N}}_2}\left( g \right) + {{\rm{O}}_2}\left( g \right) \to {\rm{N}}{{\rm{O}}_2}\left( g \right)\;\;\;\;\;\Delta H^\circ = 33.2\;{\rm{kJ}}$ …… (1)

$\frac{1}{2}{{\rm{N}}_2}\left( g \right) + \frac{1}{2}{{\rm{O}}_2}\left( g \right) \to {\rm{NO}}\left( g \right)\;\;\;\;\;\Delta H^\circ = 90.2\;{\rm{kJ}}$ …… (2)

Equation number 2 is reversed and multiplied by 2 to get,

$2{\rm{NO}}\left( g \right) \to {{\rm{N}}_2}\left( g \right) + {{\rm{O}}_2}\left( g \right)\;\;\;\;\;\Delta H^\circ = - 180.4\;{\rm{kJ}}$ …… (3)

Equation number 1 is multiplied by 2 to get,

${{\rm{N}}_2}\left( g \right) + 2{{\rm{O}}_2}\left( g \right) \to 2{\rm{N}}{{\rm{O}}_2}\left( g \right)\;\;\;\;\;\Delta H^\circ = 66.4\;{\rm{kJ}}$ …… (5)

Adding equation number 3 and 5, we get

$\begin{array}{l}\\\;\;\;\;\;\;\;\;\;2{\rm{NO}}\left( g \right) \to \cancel{{{{\rm{N}}_2}}}\left( g \right) + {{\rm{O}}_2}\left( g \right)\;\;\;\Delta H^\circ = - 180.4\;{\rm{kJ}}\\\\\frac{{\cancel{{\;{{\rm{N}}_2}}}\left( g \right) + 2{{\rm{O}}_2}\left( g \right) \to 2{\rm{N}}{{\rm{O}}_2}\left( g \right)\;\;\;\;\;\;\;\Delta H^\circ = 66.4\;{\rm{kJ}}}}{{2{\rm{NO}}\left( g \right) + {{\rm{O}}_2}\left( g \right) \to 2{\rm{N}}{{\rm{O}}_2}\left( g \right)\;\;\;\;\;\;\;\;\Delta H = - 114\;{\rm{kJ}}\;\;\;}}\\\end{array}$

Part B

The given reactions are as follows:

${{\rm{B}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\left( s \right) + 3{{\rm{H}}_2}{\rm{O}}\left( g \right) \to 3{{\rm{O}}_2}\left( g \right) + {{\rm{B}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}\;\left( g \right)\;\;\;\;\Delta H^\circ = 2035\;{\rm{kJ}}$ …… (6)

$2{\rm{B}}\left( s \right) + 3{{\rm{H}}_2}\left( g \right) \to {{\rm{B}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}\left( g \right)\;\;\;\;\;\Delta H^\circ = 36\;{\rm{kJ}}$ …… (7)

${{\rm{H}}_2}\left( g \right) + \frac{1}{2}{{\rm{O}}_2}\left( g \right) \to {{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right)\;\;\;\;\;\Delta H^\circ = - 285\;{\rm{kJ}}$ …… (8)

${{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right)\; \to {{\rm{H}}_{\rm{2}}}{\rm{O}}\left( g \right)\;\;\;\;\;\Delta H^\circ = 44\;{\rm{kJ}}$ …… (9)

Equation number 7 is multiplied by 2 to get,

$4{\rm{B}}\left( s \right) + 6{{\rm{H}}_2}\left( g \right) \to 2{{\rm{B}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}\left( g \right)\;\;\;\;\;\Delta H^\circ = 72\;{\rm{kJ}}$ …… (10)

Equation number 6 is multiplied by 2 and reversed to get,

${\rm{6}}{{\rm{O}}_2}\left( g \right) + 2{{\rm{B}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}\;\left( g \right) \to \;2{{\rm{B}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\left( s \right) + 6{{\rm{H}}_2}{\rm{O}}\left( g \right)\;\;\;\Delta H^\circ = - 4070\;{\rm{kJ}}$ …… (12)

Equation number 8 is multiplied by 6 and reversed to get,

${\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right)\; \to 6{{\rm{H}}_2}\left( g \right) + 3{{\rm{O}}_2}\left( g \right)\;\;\;\;\Delta H^\circ = 1710\;{\rm{kJ}}$ …… (13)

Equation number 9 is multiplied by 6 and reversed to get,

${\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( g \right)\; \to 6{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right)\;\;\;\;\;\Delta H^\circ = - 264\;{\rm{kJ}}$ …… (14)

Adding equation number 10, 12, 13 and 14, we get

$\begin{array}{l}\\\;\;\;\;\;\;\;\;\;4{\rm{B}}\left( s \right) + \cancel{{6{{\rm{H}}_2}}}\left( g \right) \to \cancel{{2{{\rm{B}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}}}\left( g \right)\;\;\;\;\;\Delta H^\circ = 72\;{\rm{kJ}}\\\\\;\;\;\;{\rm{6}}{{\rm{O}}_2}\left( g \right) + \cancel{{2{{\rm{B}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}}}\;\left( g \right) \to \;2{{\rm{B}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\left( s \right) + \cancel{{6{{\rm{H}}_2}{\rm{O}}}}\left( g \right)\;\;\;\Delta H^\circ = - 4070\;{\rm{kJ}}\\\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\cancel{{{\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\left( l \right)\; \to \cancel{{6{{\rm{H}}_2}}}\left( g \right) + 3{{\rm{O}}_2}\left( g \right)\;\;\;\;\Delta H^\circ = 1710\;{\rm{kJ}}\\\\\;\;\;\;\;\;\;\;\;\frac{{\cancel{{{\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\left( g \right)\; \to \cancel{{6{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\left( l \right)\;\;\;\;\;\Delta H^\circ = - 264\;{\rm{kJ}}}}{{\;\;4{\rm{B}}\left( s \right) + 3{{\rm{O}}_2}\left( g \right) \to 2{{\rm{B}}_2}{{\rm{O}}_3}\left( s \right)\;\;\;\;\;\;\;\;\;\;\;\;\Delta H = - 2552\;{\rm{kJ}}\;\;\;}}\\\end{array}$

Ans: Part A

The enthalpy of reaction is $\Delta H = - 114\;{\rm{kJ}}$ .