For the equilibrium- N2O4- 2NO2- G0N2O4298 - 100 kJ-mole and G0NO2298 - 50 kJ-mole- When 5 mole - lit is each taken- the value of - G for the reaction-

The heat of formation of CO $_{(g)}$ and CO $_{2 (g)}$ are $Δ$ H =- 110 and $Δ$ H = -393 kJ.mol $^{- 1}$ respectively. What is the heat of reaction ( $Δ$ H ) (in kJ.mol $^{- 1}$ ) for the following reaction $C O + \frac{1}{2} O_{2} \to C O_{2}$ ?

Q. The

Δ H_{f}^{o}

for

C O_{2}

(g),

C O

(g) and

H_{2} O (l)

are -393.5, -110.5 and -241.8 kJ.mol

^{- 1}

respectively. The standard enthalpy change ( in kJ) for the reaction

C O_{2} (g) + H_{2} (g) \to C O (g) + H_{2} O (l)

is:

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For the equilibrium, N2O4⇋2NO2;(G0N2O4)298 = 100 kJ/mole and (G0NO2)298 = 50 kJ/mole. When 5 mole / lit is each taken, the value of ΔG for the reaction.

The correct option is A -50 kJ.mol−1N2O4(g)⟶2NO2(q)n=5 moles.Temperature =298K, Pressure =20 barΔGoN2O4=100 kJ, ΔGoNO2=50kJΔG=ΔGNO2−ΔGN2O4=50−100=−50 kJ/mol

For the equilibrium, $N_{2} O_{4} ⇋ 2 N O_{2}; (G_{N_{2} O_{4}}^{0})_{298}$ = 100 kJ/mole and $(G_{N O_{2}}^{0})_{298}$ = 50 kJ/mole. When 5 mole / lit is each taken, the value of $Δ G$ for the reaction.

The correct option is A -50 $k J . m o l^{- 1}$
$N_{2} O_{4_{(g)}} ⟶ 2 N O_{2 (q)}$

$n = 5$ $m o l e s$ .

Temperature $= 298 K$ , Pressure $= 20 b a r$

$Δ G_{N_{2} O_{4}}^{o} = 100 k J,$ $Δ G_{N O_{2}}^{o} = 50 k J$

$Δ G = Δ G_{N O_{2}} - Δ G_{N_{2} O_{4}} = 50 - 100 = - 50 k J / m o l$