Question #8f2d8

1 Answer
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.

#2H_(2(g)) + O_(2(g)) -> 2H_2O_((l))#

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

#DeltaH_f^(0) = "-571.6 kJ"/"2 moles" = "-285.8 kJ/mol"#

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

#-"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

#DeltaH_f^(0) = color(green)(-"15.86 kJ/kg")#