Problem 6

In a hypothetical universe, an unknown amount of diborane reacts:

\ce{B2H6(g) + 6H2O(l) -> 2H3BO3(s) + 6H2(g)}

. The obtained

\ce{H3BO3(s)}

is completely sublimed at 300 K; the necessary energy is supplied as work from one cycle of an ideal heat engine in which one mole of monatomic perfect gas undergoes: A→B isothermal reversible expansion absorbing

q_H = 250

J at

T_H = 1000

K; B→D reversible adiabatic expansion; D→C isothermal reversible compression at

T_C = 300

K releasing

q_C

; C→A reversible adiabatic compression, with

q_H/q_C = T_H/T_C

. Reaction enthalpies at 300 K (kJ mol

^{-1}

): (1)

\ce{B2H6(g) + 6Cl2 -> 2BCl3(g) + 6HCl(g)}

\Delta_r H(1) = -1326

; (2)

\ce{BCl3(g) + 3H2O(l) -> H3BO3(g) + 3HCl(g)}

\Delta_r H(2) = -112.5

; (3)

\ce{B2H6(g) + 6H2O(l) -> 2H3BO3(s) + 6H2(g)}

\Delta_r H(3) = -493.4

; (4)

\ce{1/2 H2(g) + 1/2 Cl2(g) -> HCl(g)}

\Delta_r H(4) = -92.3

. (a) Calculate the molar enthalpy of sublimation of

\ce{H3BO3}

at 300 K. (b) Calculate

\Delta_r U

of reactions (2) and (4). (c) Calculate

|w|

|q_C|

and the efficiency of the engine. (d) Calculate the amount of

\ce{H2(g)}

produced per engine cycle.

Step 1 of 4: Sublimation enthalpy by Hess's law

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2\,\Delta_{\text{sub}}H(\ce{H3BO3}) = \big[\Delta_r H(1) + 2\Delta_r H(2) - 12\Delta_r H(4)\big] - \Delta_r H(3) = -443.4 + 493.4 = +50\ \text{kJ} \Rightarrow \Delta_{\text{sub}}H = +25\ \text{kJ mol}^{-1}

Analysis

Reaction (3) with $\ce{H3BO3(g)}$ instead of (s) equals (1) + 2(2) − 12(4): $-1326 - 225 + 1107.6 = -443.4$ kJ. The difference with reaction (3), $-443.4-(-493.4) = +50$ kJ, is the energy needed to sublime 2 mol of $\ce{H3BO3}$ , so $\Delta_{\text{sub}}H = +25$ kJ mol $^{-1}$ .