Question

Question Quartz,${\mathbf{SiO}}_{\mathbf{2}\mathbf{}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$, does not spontaneously decompose to silicon and oxygen at $\mathbf{25}\mathbf{°}\mathbf{}\mathbf{C}\mathbf{,}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{0}$atm in the reaction ${\mathbf{\text{SiO}}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }\mathbf{\text{Si}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$even though the standard entropy change of the reaction is large and positive $\mathbf{(}\mathbf{∆}\mathbf{S}\mathbf{°}\mathbf{}\mathbf{=}\mathbf{+}\mathbf{182}\mathbf{.}\mathbf{02}\mathbf{}\mathbf{J}\mathbf{}{\mathbf{K}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{)}$. Explain.

Short answer

It can be concluded that the reaction is definitely not spontaneous by the very positive value of the change in Gibbs energy.

Step-by-step solution

Given data

The value of entropies is. $∆\mathrm{S}°=182.02{\mathrm{JK}}^{-1}$.

The temperature is $T\mathit{}\mathit{=}\mathit{298}K$.

Concept of adiabatic process

At constant temperature and pressure, the Gibbs free energy is a thermodynamic potential that can be used to compute the maximum work that a thermodynamically closed system can perform.

Calculation of change of enthalpy

The change of enthalpy can be calculated by the formula:
$∆\mathrm{H}=\underset{}{\sum ∆\mathrm{H}°{}_{\mathrm{products}}}-\sum ∆\mathrm{H}°\mathrm{reactants}$

Where, localid="1663394907685" $\underset{}{\sum ∆\mathrm{H}{°}_{\mathrm{products}}}$is change of enthalpy of product and $\underset{}{\sum ∆\mathrm{H}{°}_{\mathrm{reactants}}}$is change of enthalpy of recant. The enthalpies of pure molecules are 0. From Appendix D: $∆\mathrm{H}°({\mathrm{SiO}}_2)=-910\mathrm{kJ}/\mathrm{mol}$

Puts the value of given data in the above equation.

$$\begin{gathered} \Delta H=0+0-(-910) \\ =910\text{kJ}/\text{mol} \end{gathered}$$

Calculation of Gibbs free energy

The Gibbs free energy is calculated with the help of the formula:
$∆\mathrm{G}=∆\mathrm{H}-\mathit{}T∆S$
Where, $∆\mathrm{G}$is Gibbs free energy,$∆\mathrm{H}$is the change in enthalpy.

Put the value of the given data in the above equation.
$$\begin{gathered} ∆\mathrm{G}=910\times {10}^3-298\times 182.02 \\ ∆\mathrm{G}=855.76\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

It can be concluded that the reaction is definitely not spontaneous by the very positive value of the change in Gibbs energy.