A sample of solid azulene (C10H8) that weighs 0.4925 g is burned in an excess of

Question

A sample of solid azulene (C10H8) that weighs 0.4925 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 2.150 °C. The heat capacity of the calorimeter and its contents is known to be 9.455×103 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 azulene to CO2(g) and H2O().

kJ mol-1

(c) Calculate the standard enthalpy of formation per mole of azulene, 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

a) The balanced equation is

C10H8 (s) + 12O2 (g) ---------------> 10 CO2(g) + 4H2O(l)

b) The heat given by sample = heat absorbed by the calorimeter.

= heat capacity x rise in temperature

= 9.455x103 J /K x 2.150K

= 2.0328x104 J

The heat given by 0.4925g of sample = -2.0328x104 J

one mole (128g/mol) of azulene gives

heat = 128g/mol x - 2.0328x104 J / 0.4925 g

= -528.328 x104 J/mol

b) Thus enthalpy of combustion azulene = -528.328 x104 J/mol (assuming H = E)

c) calculating enthalpy of formation of azulene

we know heat of reacction = sum of enthalpies of formation of products - sum of heat of formation of reactants

-528.328 x104 J/mol = [10(-393.51) +4 (-285.83) ] -[ delta Hf of azulene + 0]

Thus delta H f of azulene = -5278.202kJ /mol

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