# What is the reaction between ammonia and bleach that makes it unsafe to mix the two?

## I think bleach has strong base in it, and I know ammonia can either be an acid or a base, but I don't really have much of an idea of what the reaction between the two is. Does it form some gas?

Jun 15, 2018

The most likely reaction is to form chloramine:

$\text{NH"_3(aq) + "NaOCl"(aq) -> "NH"_2"Cl"(g) + "NaOH} \left(a q\right)$

It does look rather awkward, as we might see this and think, "but we tried to react weak base with weak base, didn't we?". [Ammonia is typically a base, and ${\text{OCl}}^{-}$ is the weak conjugate base of the weak acid $\text{HOCl}$.]

However, this reputable source states that the above is a "well-known process" for when ${\text{NH}}_{3}$ reacts with $\text{HOCl}$ to eventually form ${\text{NCl}}_{3}$ in a three-step mechanism.

Note that chloramine, $\text{NH"_2"Cl}$, supposedly may cause tumors and genetic mutation, and isn't particularly stable, which is why this reaction continues further.

Based on that source, we can estimate the thermodynamic likelihood of this being a possible reaction. The remaining $\Delta {H}_{f}^{\circ}$ were found from Levine's Physical Chemistry (${\text{OH}}^{-}$), here (${\text{OCl}}^{-}$), and here (${\text{NH}}_{3}$).

$\text{NH"_2"Cl": DeltaH_f^@ ~~ "13.74 kcal/mol" = "57.49 kJ/mol}$

${\text{OH}}^{-} \left(a q\right)$, $\Delta {H}_{f}^{\circ} = - \text{229.994 kJ/mol}$

${\text{OCl}}^{-} \left(a q\right)$, $\Delta {H}_{f}^{\circ} = - \text{118.34 kJ/mol}$

${\text{NH}}_{3} \left(g\right)$, $\Delta {H}_{f}^{\circ} = - \text{45.90 kJ/mol}$

So this is for the reaction:

${\text{NH"_3(aq) + "OCl"^(-)(aq)-> "NH"_2"Cl"(g) + "OH}}^{-} \left(a q\right)$

This gives:

DeltaH_(rxn)^@ = ["57.49 kJ/mol" + (-"229.994 kJ/mol")] - [-"45.90 kJ/mol" + (-"118.34 kJ/mol")]

$= - \text{8.26 kJ/mol}$

This reaction forms a gas from aqueous reactants, so the entropy change must be positive.

Hence, the change in Gibbs' free energy is:

$\Delta {G}_{r x n}^{\circ} = \Delta {H}_{r x n}^{\circ} - T \Delta {S}_{r x n}^{\circ}$

$= \left(-\right) - \left(+\right) \left(+\right)$

$= \left(-\right)$

and this reaction is therefore spontaneous at $\text{300 K}$.