The following information is required for this question: $\Delta H_f^\circ(\mathrm{P_4O_{10}(s)})=-3012\,\mathrm{kJ\,mol^{-1}}$, $\Delta H_f^\circ(\mathrm{H_2O(l)})=-286\,\mathrm{kJ\,mol^{-1}}$, $\Delta H_f^\circ(\mathrm{H_3PO_4(s)})=-1279\,\mathrm{kJ\,mol^{-1}}$. Find $\Delta H^\circ$ for the reaction shown. $\mathrm{P_4O_{10}(s) + 6H_2O(l) \rightarrow 4H_3PO_4(s)}$
- A$-9844\,\mathrm{kJ\,mol^{-1}}$
- B$-388\,\mathrm{kJ\,mol^{-1}}$
- C$-97\,\mathrm{kJ\,mol^{-1}}$
- D$+2019\,\mathrm{kJ\,mol^{-1}}$