Question #67549

You are provided with the thermochemical equation that describes the dissociation of sodium hydroxide into its constituent ions when dissolved in water.

#"NaOH"_ ((s)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-), " " DeltaH_"rxn" = - "46.80 kJ mol"^(-1)#

In this case, the enthalpy change of reaction is essentially the enthalpy change of solution of sodium hydroxide, i.e. the change in enthalpy that occurs when sodium hydroxide is dissolved in water at constant pressure to get an infinite dilution.

The enthalpy change of solution tells you that when #1# mole of sodium hydroxide dissolves in water, #"46.80 kJ"# of heat are being given off.

Notice that you have

#DeltaH_"rxn" = - "46.80 kJ mol"^(-1)#

The minus sign is used here to show that heat is being given off.

So, what would happen if you were to dissolve #4# moles of sodium hydroxide in water?

Since you know that #1# mole of sodium hydroxide releases #"46.80 kJ"# of heat when dissolved in water, you can say that using #4# times as many moles will cause

#4 color(red)(cancel(color(black)("moles NaOH"))) * overbrace("46.80 kJ"/(1color(red)(cancel(color(black)("mole NaOH")))))^(color(blue)(= DeltaH_"rxn")) = "187.2 kJ"#

of heat to be released. This implies that the enthalpy change of reaction when #4# moles of sodium hydroxide are dissolved in water will be

#DeltaH_"rxn 4 moles" = - "187.2 kJ"#

Once again, we use the minus sign to show that heat is being given off.

So regardless on how many grams of sodium hydroxide you have, you can say that using #4# times as much sodium hydroxide will cause the reaction to give off #4# times as much heat.