Question #886b4

1 Answer
Sep 3, 2015

Answer:

The answer is (a) reactions (i) and (ii)

Explanation:

FULL QUESTION

Consider the following chemical reactions (note that all of the aqueous species will be fully dissociated into ions).

Which two reactions should have the same value of #DeltaH#?

(a) (i) and (ii) (b) (i) and (iii) (c) (ii) and (iii) (d) (ii) and (iv) (e) (iii) and (iv)

#HCl_((aq)) + NaOH_((aq))-> NaCl_((aq)) + H_2O_((l))" "color(blue)((1))#

#HNO_(3(aq)) + KOH_((aq)) -> KNO_(3(aq)) + H_2O_((l))" "color(blue)((2))#

#HCl_((aq)) + NaOH_((s)) -> NaCl_((aq)) + H_2O_((l))" "color(blue)((3))#

#HNO3_((aq)) + KOH_((s)) -> KNO_(3(aq)) + H_2O_((l))" "color(blue)((4))#

#stackrel("------------------------------------------------------------------------------------------------------------------")#

All four reactions are neutralization reactions that take place between strong acids and strong bases. The only difference between them is the state of the strong base.

The idea here is that when you mix an aqueous solution that contains a strong acid and an aqueous solution that contains a strong base, the only reaction that takes place is the neutralization reaction.

The enthalpy change of reaction will be determined exclusively by the enthalpy change of neutralization, which is approximately equal for any strong acid - strong base pair.

On the other hand, when you add a solid strong base to an aqueous solution that contains a strong acid, two reactions take place.

First, the strong base will dissociate into cations and anions, then the neutralization reaction will take place.

This means that the enthalpy change of reaction will no longer be determined exclusively by the enthalpy change of neutralization, it will have to account for the enthalpy change of solution as well.

The dissociation in aqueous solution for both sodium hydroxide and potassium hydroxide is exothermic, meaning that it releases heat. However, there's a difference between them.

Sodium hydroxide's enthalpy change of solution (at #25^@"C"#) is equal to #-"44.51 kJ/mol"#, while potassium hydroxide's enthalpy change of solution is equal to #-"57.61 kJ/mol"#.

This means that reactions #color(blue)((3))# and #color(blue)((4))# will not have the same overall enthalpy change of reaction.

This means that the only two reactions that will have the same enthalpy change of reaction will be #color(blue)((1))# and #color(blue)((2))#, since they both feature the neutralization reaction betwween strong acids and strong bases.