Answer to Question #86918 in General Chemistry for mike

Question

A sample of solid dimethyl oxalate (C4H6O4) that weighs 1.431 g is burned in an excess of oxygen to CO2(g) and H2O() in a constant-volume calorimeter at 25.00 °C. The temperature rise is observed to be 2.140 °C. The heat capacity of the calorimeter and its contents is known to be 9.474×103 J K-1.

(a) Write and balance the chemical equation for the combustion reaction. Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.


(b) Assuming that H° is approximately equal to E, calculate the standard enthalpy change for the combustion of 1.000 mol of dimethyl oxalate to CO2(g) and H2O().

_________ kJ mol-1

(c) Calculate the standard enthalpy of formation per mole of dimethyl oxalate, using the following for the standard enthalpies of formation of CO2(g) and H2O().
Hf° H2O () = -285.83 kJ mol-1 ; Hf° CO2(g) = -393.51 kJ mol-1

________ kJ mol-1

Answer 1

(a)

$$2C_2H_6O_4 + 7 O_2 \rightarrow 8 CO_2 + 6 H_2O$$

(b)

$$q = c_{cal}\times \Delta T = 9.474\times 10^3 \frac{J}{K} \times 2.140 K = 20274.36 J = 20.274 kJ$$

$$n(C_4H_6O_4) = \frac{m}{M} = \frac{1.431 g}{118.089\frac{g}{mol}} = 0.0121 mol$$

$$\Delta H_c^0 = -\frac{q}{n} = -\frac{20.274 kJ}{0.0121 mol} = -1676 \frac{kJ}{mol}$$

(c)

$$C_2H_6O_4 + 3.5 O_2 \rightarrow 4 CO_2 + 3 H_2O$$

$$\Delta H_c^0 = \sum\Delta H_f^0 (products) - \sum\Delta H_f^0 (reactants)$$ $$\Delta H_c^0 = (4\times \Delta H_f^0 (CO_2) + 3 \times \Delta H_f^0 (H_2O)) - (3.5\times \Delta h_f^0 (O_2) +\Delta H_f^0 (C_4H_6O_4))$$

$$-1676 \frac{kJ}{mol} = (4\times (-393.51 \frac{kJ}{mol})+ 3\times (-285.83 \frac{kJ}{mol})) - (3.5\times 0 \frac{kJ}{mol} + \Delta H_f^0 (C_4H_6O_4))$$

$$\Delta H_f^0 (C_4H_6O_4) = -755.53 \frac{kJ}{mol}$$