Question

Cassiterite when reduced with coke Tin is obtained. What is the minimum temperature needed for the reduction of cassiterite would take place?

At 298 K ${∆}_{\mathrm{f}}\mathrm{H}°=\left({\mathrm{SnO}}_2\left(\mathrm{S}\right)\right)=-580.0{\mathrm{kJmol}}^{-1},{∆}_{\mathrm{f}}\mathrm{H}°\left({\mathrm{CO}}_2\left(\mathrm{g}\right)\right)=-394.0{\mathrm{kJmol}}^{-1}$

$\mathrm{S}°\left({\mathrm{SnO}}_2\left(\mathrm{s}\right)\right)=56.0{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1},\mathrm{S}°\left(\mathrm{Sn}\left(\mathrm{s}\right)\right)=52.0{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$

$\mathrm{S}°\left(\mathrm{C}\left(\mathrm{s}\right)\right)=6.0{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1},°\left({\mathrm{CO}}_2\left(\mathrm{g}\right)\right)=210.0{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$

Assume that enthalpies and entropies are temperature independent.

Answer 1

Given:

$$\begin{gathered} {\mathrm{SnO}}_2\left(\mathrm{s}\right)+\mathrm{C}\left(\mathrm{s}\right)\to \mathrm{Sn}\left(\mathrm{l}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right) \\ \mathrm{Tin}\mathrm{Oxide}\mathrm{Carbon}\mathrm{Tin}\mathrm{Carbon}\mathrm{Dioxide} \end{gathered}$$

$∆\mathrm{H}=∆{\mathrm{H}}_{\mathrm{f}}\left(\mathrm{p}\right)-∆{\mathrm{H}}_{\mathrm{f}}\left(\mathrm{r}\right)$

where, $∆\mathrm{H}$ is enthalpy,$∆{\mathrm{H}}_{\mathrm{f}}\left(\mathrm{p}\right)$ is enthalpy at constant pressure, and $∆{\mathrm{H}}_{\mathrm{f}}\left(\mathrm{r}\right)$ is the enthalpy of formation.

putting the values in above expression:

$\begin{array}{rcl}∆\mathrm{H}& =& -394-\left(581\right)\\ & =& 187{\mathrm{KJmol}}^{-1}\end{array}$

Expression used for the calculating the entropy:

$∆\mathrm{S}={\mathrm{S}}_{\mathrm{p}}-{\mathrm{S}}_{\mathrm{r}}$

$\text{∆S}$ is the change in entropy, ${\mathrm{S}}_{\mathrm{p}}$ enthalpy at constant pressure, and ${\mathrm{S}}_{\mathrm{r}}$ is the entropy of formation.

putting values in the above expression:

$\begin{array}{rcl}∆S& =& \left(52+210\right)-\left(56+6\right)\\ & =& 202-62\\ & =& 200\end{array}$

Expression used to calculate the minimum temperature:

$\therefore ∆\mathrm{G}=∆\mathrm{H}-\mathrm{T}∆\mathrm{S}$

$∆\mathrm{G}$ is the free energy,$∆\mathrm{H}$ is the change in enthalpy and $∆\mathrm{S}$ is change in entropy.

$$\begin{gathered} 0=18700-\mathrm{T}\times 200 \\ 200\mathrm{T}=18700 \\ \mathrm{T}=935\mathrm{K} \end{gathered}$$

Therefore, $935\mathrm{K}$ is the minimum temperature needed for the reduction of cassiterite would take place.