Question

Be sure to answer all parts. Sulfur dioxide is released in the combustion of coal. Scrubbers use lime slurries of calcium hydroxide to remove the SO2 from the flue gases. Write the balanced equation for the reaction between solid calcium hydroxide and SO2. Include the states of all reactants and products in your equation. Now, calculate the ΔS o at 298 K [S o of CaSO3(s) = 101.4 J/mol K].

Answer 1

Answer: The balanced chemical equation is written below and $\Delta S^o$ for the reaction is -160.6 J/K

Explanation:

When calcium hydroxide reacts with sulfur dioxide, it leads to the formation of calcium sulfate and water molecule.

The chemical equation for the reaction of calcium hydroxide and sulfur dioxide follows:

$Ca(OH)_2(s)+SO_2(g)\rightarrow CaSO_3(s)+H_2O(l)$

To calculate the entropy change of the reaction, we use the equation:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{products}]-\sum [n\times \Delta S^o_{reactants}]$

For the given reaction:

$\Delta S^o_{rxn}=[(1\times \Delta S^o_{CaSO_3(s)})+(1\times \Delta S^o_{H_2O(l)})]-[(1\times \Delta S^o_{Ca(OH)_2(s)})+(1\times \Delta S^o_{SO_2(g)})]$

Taking the standard entropy change values:

$\Delta S^o_{CaSO_3(s)}=101.4Jmol^{-1}K^{-1}\\\Delta S^o_{H_2O(l)}=69.9Jmol^{-1}K^{-1}\\\Delta S^o_{Ca(OH)_2(s)}=83.4Jmol^{-1}K^{-1}\\\Delta S^o_{SO_2(g)}=248.5Jmol^{-1}K^{-1}$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(1\times (101.4))+(1\times (69.9))]-[(1\times (83.4))+(1\times (248.5))]\\\\\Delta S^o_{rxn}=-160.6J/K$

Hence, the balanced chemical equation is written above and $\Delta S^o$ for the reaction is -160.6 J/K