Problem 1

A compound Q (molar mass 122.0 g mol

^{-1}

) consists of carbon, hydrogen and oxygen. PART A. The standard enthalpies of formation of

\ce{CO2(g)}

and

\ce{H2O(l)}

25.00\ ^\circ\mathrm{C}

are

-393.51

and

-285.83

kJ mol

^{-1}

;

R = 8.314

J K

^{-1}

mol

^{-1}

(atomic masses: H = 1.0, C = 12.0, O = 16.0). A 0.6000 g sample of solid Q is combusted in excess oxygen in a bomb calorimeter initially containing 710.0 g of water at

25.000\ ^\circ\mathrm{C}

. After reaction the temperature is

27.250\ ^\circ\mathrm{C}

and 1.5144 g of

\ce{CO2(g)}

and 0.2656 g of

\ce{H2O(l)}

are produced. (1.1) Determine the molecular formula of Q and write a balanced equation, with states, for its combustion. (1.2) Given the specific heat of water 4.184 J g

^{-1}

^{-1}

and the internal energy change of the reaction

\Delta U^\circ = -3079

kJ mol

^{-1}

, calculate the heat capacity of the calorimeter (excluding the water). (1.3) Calculate the standard enthalpy of formation

\Delta H_f^\circ

of Q. PART B. The distribution of Q between benzene and water at

6\ ^\circ\mathrm{C}

gave the equilibrium concentrations

(c_B, c_W)

in mol dm

^{-3}

: (0.0118, 0.00281), (0.0478, 0.00566), (0.0981, 0.00812), (0.156, 0.0102). Assume only one species of Q exists in benzene and that Q is a monomer in water. (1.4) Show by calculation whether Q is a monomer or a dimer in benzene. (1.5) For an ideal dilute solution,

\Delta T_f = R(T_f^0)^2 X_s/\Delta H_f

; the molar mass of benzene is 78.0 g mol

^{-1}

, pure benzene freezes at

5.40\ ^\circ\mathrm{C}

at 1 atm and

\Delta H_f = 9.89

kJ mol

^{-1}

. Calculate the freezing point of a solution of 0.244 g of Q in 5.85 g of benzene.

Step 3 of 5: Standard enthalpy of formation

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\Delta H^\circ = \Delta U^\circ + RT\,\Delta n_g = -3079 + (8.314\times10^{-3})(298)(-0.5) \approx -3080\ \text{kJ mol}^{-1};\quad \Delta H_f^\circ(\ce{Q}) = 7(-393.51) + 3(-285.83) - (-3080) = -532\ \text{kJ mol}^{-1}

Analysis

With $\Delta n_g = 7 - 15/2 = -0.5$ mol of gas, $\Delta H^\circ = \Delta U^\circ + RT\Delta n_g \approx -3080$ kJ mol $^{-1}$ . Hess's law applied to the combustion, $\Delta H^\circ = 7\Delta H_f^\circ(\ce{CO2}) + 3\Delta H_f^\circ(\ce{H2O}) - \Delta H_f^\circ(\ce{Q})$ , yields $\Delta H_f^\circ(\ce{Q}) = -532$ kJ mol $^{-1}$ .