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 4 of 4: Hydrogen produced per cycle

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n(\ce{H3BO3}) = \dfrac{175}{25000} = 7.0\times 10^{-3}\ \text{mol},\quad n(\ce{H2}) = 3\,n(\ce{H3BO3}) = 2.1\times 10^{-2}\ \text{mol}

Analysis

The 175 J of work sublimes $175/25000 = 7.0\times 10^{-3}$ mol of $\ce{H3BO3}$ . Since the reaction makes 6 mol H2 per 2 mol $\ce{H3BO3}$ , $n(\ce{H2}) = 3\times 7.0\times 10^{-3} = 2.1\times 10^{-2}$ mol per cycle.

Common pitfall. The work sublimes H3BO3, it does not drive the B2H6 reaction directly; the 3:1 H2:H3BO3 ratio then converts sublimed moles into H2 moles.