Answer:
6.29 J/(g*K)
Explanation:
Hi, you haven't provided the equation that we need for the calculation of entropy change. However, the equation below will be used for the calculation:
[tex]2CO_{(g)} + O_{2(g)} ---2CO_{2(g)}[/tex]
Δ[tex]s_{rxn} =[/tex] 2*Δ[tex]s_{f}(CO_{2(g)})[/tex]] - [2*Δ[tex]s_{f}(CO_{(g)})[/tex] + Δ[tex]s_{f}(O_{2(g)})[/tex]]
Δ[tex]s_{rxn}[/tex] is the standard entropy change for the reaction [J/(mol.K)]
Δ[tex]s_{f}(CO_{2(g)})[/tex] is the standard entropy of formation of carbon dioxide gas = 213.79 J/(mol.K)
Δ[tex]s_{f}(CO_{(g)})[/tex] is the standard entropy of formation of carbon monoxide gas = 197.66 J/(mol.K)
Δ[tex]s_{f}(O_{2(g)})[/tex] is the standard entropy of formation of oxygen gas = 205.03 J/(mol.K)
Therefore,
Δ[tex]s_{rxn} [/tex] = [2*213.79] - [2*197.66+205.03] = - 172.77 J/(mol*K)
From the given chemical reaction, when 2 moles of carbon monoxide gas reacts, the entropy change for the reaction -172.77 J/(mol*K). Therefore, when 2.04 moles of carbon monoxide gas reacts, the entropy change for the reaction is -172.77*2.04/2 = -176.23 J/(mol*K).
Thus, converting J/(mol*K) to J/(g*K):
molar mass of carbon monoxide is 28 g/mol.
Thus, the entropy change in J/(g*K) = -176.23/28 = 6.29 J/(g*K)
