Answer:
[tex]\Delta _RS^o=-230J/(mol*K)[/tex]
Explanation:
Hello,
In this case, for the given chemical reaction:
[tex]2 CO(g) + 2 NO(g) \rightarrow N_2(g) + 2 CO_2(g)[/tex]
The standard entropy change is computed by subtracting the products standard enthalpy and the reactants standard enthalpy considering each species stoichiometric coefficients:
[tex]\Delta _RS^o=S ^o_{N2(g)}+2S ^o_{CO2(g)}-2S ^o_{CO(g)}-2S ^o_{NO(g)}\\\\\Delta _RS^o=191.6J/(mol*K)+2*213.8J/(mol*K) -2*213.8J/(mol*K) -2*210.8J/(mol*K)\\\\\Delta _RS^o=-230J/(mol*K)[/tex]
Best regards.
Answer:
The standard entropy change for the reaction is -197.8 J/mol*K
Explanation:
Step 1: Data given
S°(CO(g)) = 197.7 J/(mol*K)
S°(CO2(g)) = 213.8 J/(mol*K)
S°(NO(g)) = 210.8 J/(mol*K)
S°(N2(g)) = 191.6 J/(mol·K)
Step 2: The balanced equation
2 CO(g) + 2 NO(g) → N2(g) + 2 CO2(g)
Step 3: Calculate ΔS°
ΔS° = ∑S°(products) - ∑S°(reactants)
ΔS° = (191.6 + 2*213.8) - (2*210.8+2*197.7) J/mol*K
ΔS° = 619.2 J/mol*K - 817.0 J/mol *K
ΔS° = -197.8 J/mol* K
The standard entropy change for the reaction is -197.8 J/mol*K
