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-

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Standard XII

Chemistry

Le Chatelier's Principle for Del N Equals to 0

For the equil...

Question

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.

-50

k J . m o l^{- 1}

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38.9

k J . m o l^{- 1}

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30.99

k J . m o l^{- 1}

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-39.9

k J . m o l^{- 1}

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Solution

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 $= 20 b a r$

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

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

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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 $= 20 b a r$

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

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