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 3 of 4: Carnot engine work and efficiency

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|q_C| = q_H\,\dfrac{T_C}{T_H} = 250\times\dfrac{300}{1000} = 75\ \text{J},\quad |w| = q_H - |q_C| = 175\ \text{J},\quad \eta = 1 - \dfrac{T_C}{T_H} = 0.70

Analysis

For a reversible cycle between two heat baths, $q_C/q_H = T_C/T_H$ , so $|q_C| = 75$ J, $|w| = 250-75 = 175$ J, and the efficiency is $\eta = |w|/q_H = 1 - T_C/T_H = 70\%$ .