Question

130. Consider the reaction:
${\mathbf{H}}_{\mathbf{2}}\mathbf{S}\left(g\right)\mathbf{+}{\mathbf{SO}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{3}\mathbf{S}\left(g\right)\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)$
for which $\mathbf{∆}\mathbf{H}$is - 233 kJ and$\mathbf{∆}\mathbf{S}$ is -424 J/K.
a. Calculate the free energy change for the reaction ($\mathbf{∆}\mathbf{G}$) at 393 K.
b. Assuming $\mathbf{∆}\mathbf{H}$and $\mathbf{∆}\mathbf{S}$do not depend on temperature, at what temperatures is this reaction spontaneous?

Short answer

1. The free energy change for the reaction($\mathbf{∆}\mathbf{G}$) at 393 K is$\mathbf{-}\mathbf{66}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{kJ}$.
2. The temperatures at which the given reaction is spontaneous at $\mathbf{T}\mathbf{<}\mathbf{550}\mathbf{K}$.

Step-by-step solution

Step 1: Introduction to the Concept

For a particular reaction, the Gibbs free energy change ($\mathbf{∆}\mathbf{G}$) is determined by the enthalpy change ($\mathbf{∆}\mathbf{H}$), entropy change ($\mathbf{∆}\mathbf{S}$), and temperature.

$\mathbf{∆}\mathbf{G}\mathbf{=}\mathbf{∆}\mathbf{H}\mathbf{-}\mathbf{T}\mathbf{∆}\mathbf{S}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\left(1\right)$

For an equilibrium reaction, $\mathbf{∆}\mathbf{G}\mathbf{=}\mathbf{0}$ and

$\mathbf{∆}\mathbf{H}\mathbf{=}\mathbf{T}\mathbf{∆}\mathbf{S}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\left(2\right)$

The sign of $\mathbf{∆}\mathbf{G}$ is determined by the total of the two variables in equation (1). Therefore, for a reaction to be spontaneous,$\mathbf{∆}\mathbf{G}\mathbf{<}\mathbf{0}$.

Step 2: Determination of  ∆G

a)

From the given,

$$\begin{gathered} ∆\mathrm{H}=-233\mathrm{kJ} \\ ∆\mathrm{S}=-424\mathrm{J}/\mathrm{K} \end{gathered}$$
${\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)+{\mathrm{SO}}_2\left(\mathrm{g}\right)\to 3\mathrm{S}\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

By equation (1),

$\mathbf{∆}\mathbf{G}\mathbf{=}\mathbf{∆}\mathbf{H}\mathbf{-}\mathbf{T}\mathbf{∆}\mathbf{S}$

$\begin{array}{rcl}∆\mathrm{G}& =& -233\mathrm{kJ}-\left(393\mathrm{K}\right)\times \left(-0.424\mathrm{kJ}/\mathrm{K}\right)\\ & =& -66.4\mathrm{kJ}\end{array}$

Step 3: Determination of Temperature

b)

We know that at equilibrium,$\mathbf{∆}\mathbf{G}\mathbf{=}\mathbf{0}$

By equation (2),

$\begin{array}{rcl}\mathbf{∆}\mathbf{H}& \mathbf{=}& \mathbf{T}\mathbf{∆}\mathbf{S}\\ \mathrm{T}& =& \frac{∆\mathrm{H}}{∆\mathrm{S}}\\ & =& \frac{-233\mathrm{kJ}}{-0.424\mathrm{kJ}/\mathrm{K}}\\ & =& 550\mathrm{K}\end{array}$

$\mathbf{∆}\mathbf{G}$ must be negative for the reaction to be spontaneous. Because both $∆\mathrm{H}$and $∆\mathrm{S}$ are negative, the reaction will spontaneously occur at temperatures below 550 K, with the$\mathbf{∆}\mathbf{H}$term dominating. If the temperature is above 550 K the $\mathbf{∆}\mathbf{S}$term will be dominating and the reaction would not be spontaneous.