At 27^o c,  for the combustion of one mole of ethane, heat of reaction at constant volume is equal to -1560 kJ/mole. Heat of reaction at constant pressure would be

$$\begin{gathered} {\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)+\frac{7}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\stackrel{ }{\to }2{\mathrm{CO}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right); \\ ∆{\mathrm{n}}_{\mathrm{g}}=2-4.5=-2.5; \\ ∆\mathrm{H}=∆\mathrm{E}+∆\mathrm{nRT}; \\ =-1560 \mathrm{kJ}-2.5(8.314\times {10}^{-3} \mathrm{kJ} {\mathrm{mol}}^{-1}{\mathrm{K}}^{-1})(300 \mathrm{K}); \\ =(-1560-6.24) \mathrm{kJ}/\mathrm{mole} \\ ∆\mathrm{H}=-1566.24 \mathrm{kJ}/\mathrm{mole} ; \end{gathered}$$