Answer to Question #86918 in General Chemistry for mike

Question #86918
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
1
Expert's answer
2019-03-26T04:18:40-0400

(a)

2C2H6O4+7O28CO2+6H2O2C_2H_6O_4 + 7 O_2 \rightarrow 8 CO_2 + 6 H_2O

(b)

q=ccal×ΔT=9.474×103JK×2.140K=20274.36J=20.274kJq = c_{cal}\times \Delta T = 9.474\times 10^3 \frac{J}{K} \times 2.140 K = 20274.36 J = 20.274 kJ

n(C4H6O4)=mM=1.431g118.089gmol=0.0121moln(C_4H_6O_4) = \frac{m}{M} = \frac{1.431 g}{118.089\frac{g}{mol}} = 0.0121 mol

ΔHc0=qn=20.274kJ0.0121mol=1676kJmol\Delta H_c^0 = -\frac{q}{n} = -\frac{20.274 kJ}{0.0121 mol} = -1676 \frac{kJ}{mol}

(c)


C2H6O4+3.5O24CO2+3H2OC_2H_6O_4 + 3.5 O_2 \rightarrow 4 CO_2 + 3 H_2O

ΔHc0=ΔHf0(products)ΔHf0(reactants)\Delta H_c^0 = \sum\Delta H_f^0 (products) - \sum\Delta H_f^0 (reactants)ΔHc0=(4×ΔHf0(CO2)+3×ΔHf0(H2O))(3.5×Δhf0(O2)+ΔHf0(C4H6O4))\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))

1676kJmol=(4×(393.51kJmol)+3×(285.83kJmol))(3.5×0kJmol+ΔHf0(C4H6O4))-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))

ΔHf0(C4H6O4)=755.53kJmol\Delta H_f^0 (C_4H_6O_4) = -755.53 \frac{kJ}{mol}


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