A sample of solid xanthone (C13H8O2) that weighs 0.6715 g is burned in an excess

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

Explanation / Answer

The balanced equation is

C13H8O2(s) + 14O2 (g) ---------------> 13CO2(g) + 4H2O(l)

heat given in combustion = heat absorbed by calorimeter

= heat capacity x rise in temperature

= 9.586x103 J/K x2.170 K

=2.08016x104 J

As the heat is given out in combution

heat given by xanthone = - 2.08016x104 J

b) Assuming delta E equal to delta H

0.6715 g of xanthone gives -2.08016x104 J of heat

1.0moles ( 196g/mol) of santhone gives

= 2.08016x104 Jx196/0.6715

= 6071.65 kJ /mol

c) For calculating enthalpy of formation of xanthone

   C13H8O2(s) + 14O2 (g) ---------------> 13CO2(g) + 4H2O(l)

heat of reaction (combustion) = sum of heats of formation of products - sum of heats formation of reactnts

6071.65 kJ = [ 13 x(-393.51) + 4x (-285.83) ] - [ delta Hf of xanthone -0]

Thus delta H f of xanthone= -12328.48kJ

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