a) There are two isomers of butane: butane (C4H10) and isobutane (methylpropane) (iso- C4H10). Determine the standard enthalpy of isomerisation Δh° of butane to isobutane in terms of the enthalpies of formation of the two isomers, hC4H10 and hiso-C4H10 .b) The lunar module ‘Eagle’ of the

Question

a) There are two isomers of butane: butane [latex](C_4H_{10})[/latex] and isobutane (methylpropane) (iso- [latex] C_4H_{10})[/latex]. Determine the standard enthalpy of isomerisation Δh° of butane to isobutane in terms of the enthalpies of formation of the two isomers, [latex]h_{C_4H_{10}}[/latex] and [latex]h_{iso-C_4H_{10}}[/latex]

b) The lunar module ‘Eagle’ of the Apollo mission was propelled using the energy released by the reaction:

[latex] H_2NN(CH_3)_2(l) + 2 N_2O_4(l) → 3 N_2(g) + 2 CO_2(g) + 4 H_2O(g)[/latex].

Determine the molar enthalpy Δh° of this exothermic reaction in terms of the enthalpies of formation of the reactants, [latex] h_{H_2NN(CH_3)_2(l)}, h_{N_2O_4(l)}[/latex] and of the products [latex] h_{N_2(g)} h_{CO_2(g)},h_{H_2O}[/latex].

c) The combustion of acetylene [latex](C_2H_2)[/latex] is described by the chemical reaction:

[latex] C_2H_2(g) + \frac{5}{2} O_2(g) → 2 CO_2(g) + H_2O(l) [/latex].

Determine the enthalpy of formation [latex]h_{C_2H_2}[/latex] of acetylene[latex] (C_2H_2)[/latex] in terms of the molar enthalpies [latex]h_{O_2(g)}, h_{CO_2(g)}, h_{H_2O(g)}[/latex], the molar enthalpy of the reaction Δh° and the vaporisation molar enthalpy of water [latex]h_{νap}[/latex].

Numerical Application:

a) [latex]h_{C_4H_{10}}[/latex] = −2, 877 kJ/mol, [latex]h_{i-C_4H_{10}}[/latex] = −2, 869 kJ/mol,
b) [latex]h_{H_2O(g)}[/latex] = −242 kJ/mol, [latex]h_{CO_2(g)}[/latex] = −394 kJ/mol, [latex]h_{N_2(g)}[/latex] = 0kJ/mol, [latex]h_{N_2O_4(l)}[/latex] = 10 kJ/mol, [latex]h_{H_2NN(CH_3)_2(l)}[/latex] = 52 kJ/mol
c) Δh° = −1, 300 kJ/mol, [latex]h_{vap}[/latex] = 44 kJ/mol, [latex]h_{O_2(g)}[/latex] = 0kJ/mol,

Accepted Answer

a) According to Hess’ law (8.53), the standard enthalpy of isomerisation Δh° of butane to isobutane is the difference between the enthalpy of formation of isobutane and the enthalpy of formation of butane,

[latex] \Delta h^\circ = h_C + \sum\limits_{A=1}^{r-1}{ u _{aA}h_A} [/latex] (8.53)

[latex] Δh° = h_{iso-C_4H_{10}} − h_{C_4H_{10}} [/latex] = 8kJ/mol.

which implies that the isomerisation is endothermic since Δh° > 0.

b) The molar enthalpy released Δh° by this exothermic reaction is obtained by applying Hess’ law (8.53),

[latex] Δh° = 3 h_{N_2(g)}+ 2 h_{CO_2(g)} + 4 h_{H_2O(g)} − h_{H_2NN(CH_3)_2(l)} − 2 h_{N_2O_4(l)} [/latex].

= −1, 828 kJ/mol.

c) The molar enthalpy released Δh° by the combustion of acetylene [latex](C_2H_2)[/latex] is obtained by applying Hess’ law (8.53),

[latex] Δh° = 2 h_{CO_2(g)} + h_{H_2O(l)} − h_{C_2H_2} − \frac {5}{2} h_{O_2}[/latex].

where the molar enthalpy of gaseous water [latex] h_{H_2O(g)}[/latex] is equal to the sum of the molar enthalpy of liquid water [latex] h_{H_2O(l)}[/latex] and the vaporisation molar enthalpy of water [latex] h_{vap}[/latex],

[latex] h_{H_2O(l)} = h_{H_2O(g)} + h_{vap}[/latex].

Thus, the enthalpy of formation [latex]h_{C_2H_2}[/latex] of acetylene [latex] (C_2H_2)[/latex] is given by,

[latex] h_{C_2H_2} = 2 h_{CO_2(g)} + h_{H_2O(g)} − \frac{5}{2} h_{O_2} − Δh° − h_{vap}[/latex] = 226 kJ/mol.

Question 8.5: a) There are two isomers of butane: butane (C4H10) and isobu...

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