# Question #8f2d8

Apr 12, 2015

To go from kJ/mol to kJ/kg you have to use water's molar mass, which expresses the mass in grams of 1 mole of water.

$2 {H}_{2 \left(g\right)} + {O}_{2 \left(g\right)} \to 2 {H}_{2} {O}_{\left(l\right)}$

Notice that you have 2 moles of water being formed by the reaction, which means that the enthalpy change of your reaction is actually -571.6 kJ. The standard enthalpy of formation per mole of water will be

$\Delta {H}_{f}^{0} = \text{-571.6 kJ"/"2 moles" = "-285.8 kJ/mol}$

Use the fact that 1 mole of water weighs 18.015 g, which means that

$- \text{285.8 kJ"/cancel("mol") * (1cancel("mole"))/(18.015cancel("g")) * (1000cancel("g"))/"1 kg" = -"15864.6 kJ/kg}$

Rounded to four sig figs and expressed in kJ, the answer will be

$\Delta {H}_{f}^{0} = \textcolor{g r e e n}{- \text{15.86 kJ/kg}}$