Question

what will be the entropy change in decomposition of caco3​

Answer 1

Explanation:

Decomposition of $$CaC{O}_{3} \rightarrow$$

$${CaC{O}_{3}}_{\left( s \right)} \longrightarrow {CaO}_{\left( s \right)} + {C{O}_{2}}_{\left( g \right)}$$

The entropy of the above reaction is increasing because a gas is being produced and the number of molecules is increased.

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Answer 2

The value of $\Delta S^o$ for the reaction is 160.5 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

The chemical equation for the decomposition of calcium carbonate follows:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

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

$\Delta S^o_{rxn}=[(1\times \Delta S^o_{(CaO)})+(1\times \Delta S^o_{(CO_2)})]-[(1\times \Delta S^o_{(CaCO_3)})]$

We are given:

$\Delta S^o_{(CaO)}=39.8J/K.mol\\\Delta S^o_{(CaCO_3)}=92.9J/K.mol\\\Delta S^o_{(CO_2)}=213.6J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(1\times (39.8))+(1\times (213.6))]-[(1\times (92.9))]\\\\\Delta S^o_{rxn}=160.5J/K$

Learn more about entropy change:

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