Question

When${\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathcal{l}\mathbf{\right)}$and${\mathbf{\text{D}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathcal{l}\mathbf{\right)}$are mixed, the following reaction occurs spontaneously:

${\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathcal{l}\mathbf{\right)}\mathbf{+}{\mathbf{\text{D}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathcal{l}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{\text{HOD}}\mathbf{\left(}\mathcal{l}\mathbf{\right)}$

There is little difference between the enthalpy of an$\mathbf{\text{O}}\mathbf{-}\mathbf{\text{H}}$bond and that of an$\mathbf{\text{O}}\mathbf{-}\mathbf{\text{D}}$bond. What is the main driving force for this reaction?

Short answer

The main driving force for this reaction is the tendency to increase entropy.

Step-by-step solution

Given information

When ${\text{H}}_2\text{O}\left(\mathcal{l}\right)$and ${\text{D}}_2\text{O}\left(\mathcal{l}\right)$ are mixed, the reaction occurs spontaneously is .

${\text{H}}_2\text{O}\left(\mathcal{l}\right)+{\text{D}}_2\text{O}\left(\mathcal{l}\right)\to 2\text{HOD}\left(\mathcal{l}\right)$

There is little difference between the enthalpy of an $\text{O}-\text{H}$bond and that of an $\text{O}-\text{D}$bond.

Concept of driving force of reactions

Enthalpyandentropyarethetwomainthermodynamicforcesofchemicalreactions

A reaction is said to be spontaneous if the change in enthalpy is negative and the

change in entropy is positive.

Theory of spontaneous reactions

The given reaction is spontaneous.

${\text{H}}_2\text{O}\left(l\right)+{\text{D}}_2\text{O}\left(l\right)\to 2\text{HOD}\left(l\right)$

Spontaneous reactions generally occur out of two reasons:

1. To reach the state of Lower energy (decreasing enthalpy)

2. To reach the state of maximum randomness (increasing entropy)

If the enthalpy doesn't change much, then the main driving force is the tendency to increase entropy.