Question

Question: (a) Use data from Appendix D to calculate the standard entropy change at$\mathbf{25}\mathbf{°}\mathbf{}\mathbf{C}$for the reaction

.${\mathbf{\text{N}}}_{\mathbf{2}}{\mathbf{\text{H}}}_{\mathbf{4}}\mathbf{\left(}\mathit{\ell }\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{\text{NO}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathit{\ell }\mathbf{\right)}$

(b) Supposes the hydrazineis$\left({\text{N}}_2{\text{H}}_4\right)$ in the gaseous, rather than liquid, state. Will the entropy change for its reaction with oxygen be higher or lower than that calculated in part (a)? (Hint: Entropies of reaction can be added when chemical equations are added, in the same way, that Hess's law allows enthalpies to be added.)

Short answer

The final value of standard entropy is given below.

a:$∆\mathrm{S}°=-116.58\mathrm{J}/\mathrm{Kmol}$

b: Lower.

Step-by-step solution

Given data

The value of standard entropies is given below:
$$\begin{gathered} \mathrm{S}°({\mathrm{N}}_2{\mathrm{H}}_4,\mathrm{I})=121.21{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°({\mathrm{O}}_2,\mathrm{g})=205.03{\mathrm{JK}}^{-1} \\ \mathrm{S}°({\mathrm{NO}}_2,\mathrm{g})=239.95{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ \mathrm{S}°({\mathrm{H}}_2\mathrm{O},\mathrm{I})=121.21{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \end{gathered}$$

Concept of adiabatic process

The standard entropy$\mathbf{(}\mathbf{∆}\mathbf{S}\mathbf{°}\mathbf{)}$is defined as$\mathbf{(}\mathbf{S}\mathbf{°}\mathbf{)}$ when all reactants and products are in their standard states. Entropies for 1 mole at 298K for a specific state, allotrope, molecular complexity, molar mass, and degree of dissolution.

Calculation of change of entropy

(a)

The change in entropy is calculated with the help of the formula:
$∆\mathrm{S}°=\underset{}{\sum \mathrm{S}{°}_{\mathrm{products}}}-\underset{}{\sum \mathrm{S}{°}_{\mathrm{reactants}}}$
Where, $\underset{}{\sum \mathrm{S}{°}_{\mathrm{products}}}$entropy of the product and$\underset{}{\sum \mathrm{S}{°}_{\mathrm{reactants}}}$ is entropy of the recant.
Put the value of given data in equation (i).
$$\begin{gathered} ∆\mathrm{S}°=2\mathrm{x}\mathrm{S}°({\mathrm{NO}}_2,\mathrm{g})+2\mathrm{x}\mathrm{S}°({\mathrm{H}}_2\mathrm{O},\mathrm{I})-\mathrm{S}°({\mathrm{N}}_2\mathrm{H},\mathrm{I})-3\mathrm{x}\mathrm{S}°({\mathrm{O}}_2,\mathrm{g}) \\ =2\mathrm{x}239.95+2\mathrm{x}69.91-121.21-3\mathrm{x}205.03 \\ =-116.85\mathrm{J}/\mathrm{Kmol} \end{gathered}$$

Explanation of standard entropy

(b)

It needs to determine if the entropy will be higher or lower than calculated in subtask (A), if${\mathrm{N}}_2{\mathrm{H}}_4$was in the gaseous state.

The standardized entropy for the gaseous state is usually higher than the one in the liquid state.

$\mathrm{S}°({\mathrm{N}}_2{\mathrm{H}}_4,\mathrm{I})<\mathrm{S}°({\mathrm{N}}_2{\mathrm{H}}_4,\mathrm{g})$

This means that the sum of the entropies of the reactants will be greater.

Since the sum of the entropies of the products stays the same, this change will result in lower entropy.