Would the enthalpy change be lower or higher when a solid NaOH reacts with an aqueous HCl, and why?

Plenty, there’s an entire field of chemistry dedicated to these: Solid-state chemistry

Pick up any introductory textbook to solid-state chemistry, and it should have numerous examples of such reactions.

HCl is a strong acid and NaOH is a strong base. The reaction between HCl and NaOH is a neutralization reaction. This is an exothermic reaction and the amount of heat produced is 57.3 KJ/mol.

Would the enthalpy change be lower or higher when a solid NaOH reacts with an aqueous HCl, and why?

Oh, for the days when English teachers insisted on clear, precise writing.

This question is ambiguous.

You ask, “lower or higher?”, and I ask back “lower or higher than what?” Are we comparing the enthalpy change of a reaction between solid NaOH and aqueous HCl to some other enthalpy change? If we’re not, then the question is worded wrong. If we are, what are we comparing it to?

I can make two guesses, but who knows. Maybe there are more ways to (mis)construe this question.

1. Would the enthalpy change be lower or higher when solid NaOH reacts with aqueous HCl than when aqueous NaOH reacts with aqueous HCl, and why?
2. Would the enthalpy change be positive or negative when solid NaOH reacts with aqueous HCl, and why?

If OP meant #2, David Min has already answered this.

If #1, the answer can be derived by breaking the overall reaction down to two steps:

[math]NaOH(s) → NaOH (aq), Delta H_{solution} is very negative[/math]

[math]NaOH (aq) + HCl (aq) → NaCl (aq) + H_2O, Delta H_{neutralization} is negative[/math]

The second reaction is, for comparison, the reaction of the two aqueous solutions. And if you’ve ever made an aqueous NaOH solution, you know the first one is highly negative. The flask gets hot enough to burn your hand if you’re holding it.

Since the topic reaction is the sum of the two individual steps, the enthalpy change is the sum of the two enthalpy changes. Clearly the reaction with solid NaOH will be far more exothermic than the all-aqueous neutralization reaction.

HCl(aq) + NaOH(aq) = NaCl + H2O

ΔG(20C) = -75.7kJ

Since ΔG (change in Gibbs Free Energy) is negative, we know that the reaction runs!

ΔH(20C) = -51.2kJ

Since ΔH (energy required) is negative, we know that the reaction gives off energy (heat) – meaning that the reaction is exothermal.

There! You’ve learned something about thermodynamics today!

I will assume that you are asking if the enthalpy change during the reaction between HCl(aq) and NaOH(s) is greater than the enthalpy change during the reaction between HCl(aq) and NaOH(aq).

The answer to this question is yes.

In the reaction between the two aqueous solutions,

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

the only exothermic process that is occurring is the reaction between OH-(aq) and H+(aq) to make water: H+(aq) + OH-(aq) → H2O(l)

In the reaction between solid NaOH and aqueous HCl there are two exothermic processes that must occur for the reaction to be complete.

a) the solid NaOH must dissolve:

NaOH(s) → NaOH(aq)

b) the H+(aq) and the OH-(aq) combine to form water:

H+(aq) + OH-(aq) → H2O(l)

According to Hess’ Law, the enthalpy change for this reaction will be equal to the sum of the enthalpy changes of reactions (a) and (b)

a) NaOH(s) → NaOH(aq)

+

b) NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

= NaOH(s) + HCl(aq) → NaCl(aq) + H2O(l)

HCl will only show acidic properties when H+ ions are formed from the dissociation of HCl molecules. This chiefly happens when HCl is dissolved in water, dissociating to produce H+ and Cl- ions. Hence HCl (aq) is an acid by the Arrhenius Theory.

In toluene, HCl remains as a molecule as it cannot dissolve in an organic compound, hence no H+ ions are present.

Hope this helps!