Question

Use standard entropies to calculate δs∘rxn for the balanced chemical equation: 2pcl3(g)+o2(g)→2pocl3(l) substance s∘(j/mol⋅k) pocl3(l) 222.5 pocl3(g) 325.5 pcl3(l) 217.1 pcl3(g) 311.8 o2(g) 205.2

Answer 1

Answer: The entropy change of reaction is -383.8

Explanation:

The chemical equation is as follows:

$2PCl_3(g)+O_2(g)\rightarrow 2POCl_3(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S_{(PCl_3(g))})]$

We are given:

$\Delta S_{(POCl_3(l))}=222.5\\\Delta S_{(O_2(g))}=205.2\\\Delta S_{(PCl_3(g))}=311.8$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (222.5))]-[(1\times 205.2)+(2\times (311.8))]$

$\Delta S^o_{rxn}=[(2\times (222.5))]-[(1\times 205.2)+(2\times (311.8))]$

The entropy change of reaction is -383.8