Q.Calculate standard enthalpy of reaction, $\mathrm{2C_2H_6(g) + 7O_2(g) \rightarrow 4\,CO_2(g) + 6\,H_2O}(l)$ Given that $\Delta_f H^0$ (CO$_2$)= -393.5 kJ mol$^{-1}$, $\Delta_f H^0$ (H$_2$O)= -285.8 kJ mol$^{-1}$ and $\Delta_f H^0$(C$_2$H$_6$) = -84.9 kJ mol$^{-1}$

The enthalpy of a chemical reaction (ΔrH) is defined as the difference between the total enthalpy of the products and the total enthalpy of the reactants, ΔrH = sum of H(products) - sum of H(reactants), with each substance's contribution weighted by its coefficient in the balanced equation. When products have more enthalpy than reactants (ΔrH positive), heat is absorbed and the reaction is ENDOTHERMIC; when reactants have more enthalpy than products (ΔrH negative), heat is released and the reaction is EXOTHERMIC. To compare different reactions fairly, enthalpies are reported for reactants and products in their STANDARD STATES -- the most stable physical form of a substance at 1 bar pressure and 298 K (or 1 M concentration for a species in solution) -- giving the STANDARD ENTHALPY OF REACTION, ΔrH0, written as part of a fully-specified THERMOCHEMICAL EQUATION (a balanced equation stating the physical states of every species and the exact ΔrH0 value, which must be scaled if the equation's coefficients are scaled, and which flips sign if the equation is reversed). Two especially useful standard enthalpies are defined this way: the STANDARD ENTHALPY OF FORMATION (ΔfH0) is the enthalpy change when one mole of a compound in its standard state forms from its constituent elements in their standard sta …