Question

What is the standard entropy change for the reaction below? 2 CO(g) + 2 NO(g) → N2(g) + 2 CO2(g) S o (CO(g)) = 197.7 J/(mol·K) S o (CO2(g)) = 213.8 J/(mol·K) S o (NO(g)) = 210.8 J/(mol·K) S o (N2(g)) = 191.6 J/(mol·K)

Answer 1

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

Answer 2

Answer:

$\Delta _RS^o=-230J/(mol*K)$

Explanation:

Hello,

In this case, for the given chemical reaction:

$2 CO(g) + 2 NO(g) \rightarrow N_2(g) + 2 CO_2(g)$

The standard entropy change is computed by subtracting the products standard enthalpy and the reactants standard enthalpy considering each species stoichiometric coefficients:

$\Delta _RS^o=S ^o_{N2(g)}+2S ^o_{CO2(g)}-2S ^o_{CO(g)}-2S ^o_{NO(g)}\\\\\Delta _RS^o=191.6J/(mol*K)+2*213.8J/(mol*K) -2*213.8J/(mol*K) -2*210.8J/(mol*K)\\\\\Delta _RS^o=-230J/(mol*K)$

Best regards.