First we need to determine the heat energy absorbed by the water:
[tex]\begin{gathered} q_{H2O}=mc\Delta T \\ q:heat\text{ }energy\text{ }absorbed(J) \\ m:mass(kg)=25.00\text{ }kg \\ c:specific\text{ }heat\text{ }capacity(4184Jkg^{-1}K^{-1}) \\ \Delta T:temperature\text{ }in\text{ }K\text{ }(273.15K+2.656\degree C=275.806K) \end{gathered}[/tex]We will determine the heat energy absorbed by the water by substituting the given values into the equation:
[tex]\begin{gathered} q_{H2O}=25.00kg\times4184Jkg^{-1}K^{-1}\times275.806K \\ q_{H2O}=28,849,307.6\text{ }J \end{gathered}[/tex]The energy released by the reaction is equivalent to the energy absorbed by the water:
[tex]q_{rxn}=-q_{H2O}[/tex]Therefore the amount of energy released by the reaction is -28,849,307.6J. We will convert this energy to kJ so we divide by 1000. Energy released is -28,849.31 kJ
We need to convert the mass of the compound to moles of the compound:
[tex]\begin{gathered} _nC_3H_4=\frac{mass}{molar\text{ }mass} \\ _nC_3H_4=\frac{6.0g}{40.06gmol^{-1}} \\ _nC_3H_4=0.15mol \end{gathered}[/tex]The enthalpy of formation for the reaction is:
[tex]\begin{gathered} \Delta H_{rxn}=\frac{q_{rxn}}{n} \\ \Delta H_{rxn}=\frac{-28,849.31kJ}{0.15mol} \\ \Delta H_{rxn}=-192,328kJmol^{-1} \\ \Delta H_{rxn}=-192,000\text{ }kJmol^{-1} \end{gathered}[/tex]Answer: The standard heat of formation or enthalpy of formation of compound X is -192,000kJ/mol.
