A0.483-g sample of nonanedioic acid (c9h16o4) is burned in a bomb calorimeter and the temperature increases from 24.70 °c to 27.20 °c. the calorimeter contains 1.01×103 g water and the bomb has a heat capacity of 867 j/°c. based on this experiment, calculate δe (kj/mol) for the combustion reaction.

The value of ΔE for the combustion reaction -4,955.76 kJ/mol.

Explanation:

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

where,

q = heat released by the reaction

= heat absorbed by the calorimeter

= heat absorbed by the water

= specific heat of calorimeter =

= specific heat of water =

= mass of water =

= change in temperature = 27.20 °C - 24.70 °C =2.5°C

Now put all the given values in the above formula, we get:

Now we have to calculate the enthalpy change for combustion reaction

where,

= enthalpy change = ?

q = heat released = -12,732.1 J = -12.7321 kJ

n = moles of nonanedioic acid

The value of ΔE for the combustion reaction -4,955.76 kJ/mol.

- lead + phosphoric acid = hydrogen + lead(ii) phosphate