Question #59a26

Start by writing down the balanced chemical equation for this reaction

#2"Na"_text((s]) + 2"H"_2"O"_text((l]) -> 2"NaOH"_text((aq]) + "H"_text(2(g]) uarr#

Now take a look at the standard enthalpy change of reaction, #DeltaH_"rxn"^@#. Notice that is has a negative value, #-"368.4 kJ"# to be precise, and that it's not given per mole.

This tells you two important things

• energy is being released by the reaction, since the enthalpy change of reaction is negative
• this is how much energy is released by the reaction when 2 moles of sodium metal react

Use sodium's molar mass to determine how many moles of sodium you get in that many grams

#2.35color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23.0color(red)(cancel(color(black)("g")))) = "0.1022 moles Na"#

So, if that much heat is being released when 2 moles of sodium react, how much heat will the reaction give off when #0.1022# moles will react?

#0.1022color(red)(cancel(color(black)("moles Na"))) * "368.4 kJ"/(2color(red)(cancel(color(black)("moles Na")))) - color(green)(+"18.8 kJ")#

Here is where you need to be careful. If you sy that heat is being released, you no longer need to add the minus sign. The sole purpose of the minus sign is to tell you that heat is being released.

How much heat is being released when two moles of sodium react? #"368.4 kJ"#

How much heat is being released when #0.1022# moles of sodium react? #"18.8 kJ"#

However, the standard enthalpy change of reaction will be negative, since it must symbolize heat lost. This means that you have

#DeltaH_"rxn"^@ = -"18.8 kJ"#

That is the change in enthalpy when #0.1022# moles of sodium react.