Answer:
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
[tex]q=[q_1+q_2][/tex]
[tex]-q=[c_1\times \Delta T+m_2\times c_2\times \Delta T][/tex]
where,
q = heat released by the reaction
[tex]q_1[/tex] = heat absorbed by the calorimeter
[tex]q_2[/tex] = heat absorbed by the water
[tex]c_1[/tex] = specific heat of calorimeter = [tex]867 J/^oC[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4.184 J/g^oC[/tex]
[tex]m_2[/tex] = mass of water = [tex]1.01\times 10^3g=1,010 g[/tex]
[tex]\Delta T[/tex] = change in temperature = 27.20 °C - 24.70 °C =2.5°C
Now put all the given values in the above formula, we get:
[tex]-q=[(867 J/^oC\times 2.5 ^oC)+(1,010\times 4.184J/g^oC\times 2.5^oC)][/tex]
[tex]q=-12,732.1 J[/tex]
Now we have to calculate the enthalpy change for combustion reaction
[tex]\Delta H=\frac{q}{m}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
q = heat released = -12,732.1 J = -12.7321 kJ
n = moles of nonanedioic acid
[tex]n=\frac{0.483 g}{188 g/mol}=0.002569 mol[/tex]
[tex]\Delta H=\frac{-12.7321 J}{0.002569 mol}=-4,955.76 kJ/mol[/tex]
The value of ΔE for the combustion reaction -4,955.76 kJ/mol.
