Question

Question: The dissolution of calcium chloride in water

${\mathbf{\text{CaCl}}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }{\mathbf{\text{Ca}}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{\text{Cl}}}^{\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$

is a spontaneous process at $\mathbf{25}\mathbf{°}\mathbf{}\mathbf{C}$even though the standard entropy change of the preceding reaction is negative $\left(∆S°=-44.7{\mathrm{JK}}^{-1}\right)$. What conclusion can you draw about the change in entropy of the surroundings in this process?

Short answer

The entropy change of the surroundings must be above $44.7{\mathrm{JK}}^{-1}$.

Step-by-step solution

Given data

The entropy of the surrounding is$∆\mathrm{S}°=-44.7{\mathrm{JK}}^{-1}$

Concept of the entropy

Entropy is a thermodynamics concept that deals with energy. Entropy is the measurement of the transition from order to disorder in the cosmos.

Condition for spontaneous

For a reaction to be spontaneous, the sum of the entropies of the system and the surroundings needs to be positive.

$∆\mathrm{S}{°}_{\mathrm{total}}=∆\mathrm{S}{°}_{\mathrm{system}}+∆\mathrm{S}°\mathrm{surroundings}>0$.

Calculation of change of entropy

The entropy change of the system can be written by $∆\mathrm{S}{°}_{\mathrm{system}}+∆\mathrm{S}{°}_{\mathrm{surroundings}}>0$.

Now, put the value of given data in above equation.

$$\begin{gathered} ∆\mathrm{S}{°}_{\mathrm{system}}+∆\mathrm{S}{°}_{\mathrm{surroundings}}>0 \\ -44.7+∆\mathrm{S}{°}_{\mathrm{surroundings}}>0 \\ ∆\mathrm{S}{°}_{\mathrm{surroundings}}>44.7{\mathrm{JK}}^{-1} \end{gathered}$$

In conclusion, the entropy change of the surroundings must be above 44.7JK^-1