Question

A sample of solid xanthone (C13H8O2) that weighs 0.6715 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 2.170 °C. The heat capacity of the calorimeter and its contents is known to be 9.586×103 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. + +

Based on this experiment: (b) Assuming that H° is approximately equal to E, calculate the standard enthalpy change for the combustion of 1.000 mol of xanthone to CO2(g) and H2O().

kJ mol-1

(c) Calculate the standard enthalpy of formation per mole of xanthone, 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) The balanced chemical equation is : C₁₃H₈O₂(s) + 14 O₂(g) 13 CO₂(g) + 4 H₂O(l)

(b) H^o = -6077.9 kJ.mol^-1

(c) H^o_f = -181.1 kJ.mol^-1

Explanation

Heat capacity of calorimeter = 9.586 x 10³ J/K

temperature rise = 2.170 ^oC = 2.170 K

Heat absorbed by calorimeter = (Heat capacity of calorimeter) * (temperature rise)

Heat absorbed by calorimeter = (9.586 x 10³ J/K) * (2.170 K)

Heat absorbed by calorimeter = 20801.62 J

Heat lost by reaction = -(Heat absorbed by calorimeter)

Heat lost by reaction = -(20801.62 J)

Heat lost by reaction = -20801.62 J

mass C₁₃H₈O₂ = 0.6715 g

moles C₁₃H₈O₂ = (mass C₁₃H₈O₂) / (molar mass C₁₃H₈O₂)

moles C₁₃H₈O₂ = (0.6715 g) / (196.2 g/mol)

moles C₁₃H₈O₂ = 0.00342 mol

standard enthalpy change, H^o = (Heat lost by reaction) / (moles C₁₃H₈O₂)

H^o = (-20801.62 J) / (0.00342 mol)

H^o = -6077896.3 J/mol

H^o = -6077.9 kJ/mol