Question

∆H1,  ∆H2, and ∆H3 are the enthalpy of the following reactions:

NaOH(s) àNa+(aq) + OH-(aq)            ∆H1

NaOH(s) + H+(aq) + Cl-(aq)àH2O(l) + Na+(aq) + Cl-(aq)    ∆H2

Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq)à H2O(l) + Na+(aq) + Cl-(aq)           ∆H3

Please write an equation that will demonstrate the relationship among ∆H1,  ∆H2, and ∆H3 based on Hess's Law.

Answer 1

Reaction 1: NaOH(s) ---> Na+(aq) + OH-(aq)

Reaction 2: NaOH(s) + H+(aq) + Cl-(aq) ---> H2O(l) + Na+(aq) + Cl-(aq)

Reaction 3: Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) ---> H2O(l) + Na+(aq) + Cl-(aq)

Reversing reaction 1 we get:

Na+(aq) + OH-(aq) ---> NaOH(s)

Adding reaction 2 to this we get:

NaOH(s) + H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) ---> NaOH(s) + H2O(l) + Na+(aq) + Cl-(aq)

Cancelling the common terms we get:

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) ---> H2O(l) + Na+(aq) + Cl-(aq)

Thus we obtain the reaction (3) after reversing reaction (1) and then adding it to (2)

So, we have:

dH₃ = -dH₁ + dH₂

Hope this helps !