Question

A 0.483-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.

Answer 1

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

Step-by-step explanation:

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

`q=[q_1+q_2]`

`-q=[c_1* \Delta T+m_2* c_2* \Delta T]`

where,

q = heat released by the reaction

`q_1` = heat absorbed by the calorimeter

`q_2` = heat absorbed by the water

`c_1` = specific heat of calorimeter =
`867 J/^oC`

`c_2` = specific heat of water =
`4.184 J/g^oC`

`m_2` = mass of water =
`1.01* 10^3g=1,010 g`

`\Delta T` = change in temperature = 27.20 °C - 24.70 °C =2.5°C

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

`-q=[(867 J/^oC* 2.5 ^oC)+(1,010* 4.184J/g^oC* 2.5^oC)]`

`q=-12,732.1 J`

Now we have to calculate the enthalpy change for combustion reaction

`\Delta H=(q)/(m)`

where,

`\Delta H` = enthalpy change = ?

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

n = moles of nonanedioic acid

`n=(0.483 g)/(188 g/mol)=0.002569 mol`

`\Delta H=(-12.7321 J)/(0.002569 mol)=-4,955.76 kJ/mol`

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