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Standard XII

Chemistry

Enthalpy of Combustion

In the follow...

Question

In the following equilibrium, $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ ,
when 5 moles of each is taken and the temperature is kept at 300 K. The total pressure was found to be 20 atm.
Given: $Δ G_{f (N_{2} O_{4})}^{\circ} = 100 k J, Δ G_{f (N O_{2})}^{\circ} = 50 k J;$ $R = 8.3 J K^{- 1} m o l^{- 1} a n d ln 10 = 2.3$
Find $Δ G (i n J)$ of the reaction at $300 K$ .

5.727

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57.27

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572.7

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5727

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Solution

The correct option is D $5727$
$χ_{N_{2} O_{4}} = \frac{5}{10} = \frac{1}{2} χ_{N O_{2}} = \frac{5}{10} = \frac{1}{2} ∴ p_{N_{2} O_{4}} = p_{N O_{2}} = \frac{1}{2} \times 20 = 10 a t m$
Reaction quotient $Q_{p} = \frac{[P_{N O_{2}}]^{2}}{P_{N_{2} O_{4}}} = \frac{100}{10} = 10 Δ G_{r e a c t i o n}^{\circ} = 2 Δ G_{f (N O_{2})}^{\circ} - Δ G_{f (N_{2} O_{4})}^{\circ} = 2 \times 50 - 100 = 0$
We know that, $Δ G = Δ G^{\circ} + R T ln Q_{p} = 0 + 8.3 \times 300 ln 10 = 8.3 \times 300 \times 2.3 = 5727 J$

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Similar questions

Q. The standard enthalpy of formation of

C F_{2} C l C F_{2} C l (g)

(Δ_{f} H_{300}^{\circ})

1 bar

and

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from its constituent elements in their standard states is

- 900 {kJ mol}^{- 1}

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R = 8.3 {J K}^{- 1} {mol}^{- 1}

)
The standard internal energy of formation

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at the same pressure and temperature is:

Q. For the equilibrium :

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

(G_{N_{2} O_{4}}^{o})_{298 K} = 100 k J m o l^{- 1} (G_{N O_{2}}^{o})_{298 K} = 50 k J m o l^{- 1}

If 5 moles of each species is taken in a 1 litre container. Then, calculate the value of

Δ G

at 298 K and predict the spontaneity of the reaction .

Q. In the following equilibrium:

N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)

when

5

mole of each are taken and temperature is kept at

298 K

, the total pressure was found to be

20

bar.
given:

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

(i) Find

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Q. The standard enthalpy of formation of

C F_{2} C l C F_{2} C l (g)

(Δ_{f} H_{300}^{\circ})

1 bar

and

300 K

from its constituent elements in their standard states is

- 900 {kJ mol}^{- 1}

. (Given,

R = 8.3 {J K}^{- 1} {mol}^{- 1}

)
The standard internal energy of formation

(Δ_{f}^{\circ} U_{300})

at the same pressure and temperature is:

Q. The molar entropy content of 1 mole of oxygen

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250 {J mol}^{- 1} K^{- 1}

. Calculate

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In the following equilibrium, N2O4(g)⇌2NO2(g), when 5 moles of each is taken and the temperature is kept at 300 K. The total pressure was found to be 20 atm. Given: ΔG∘f(N2O4)=100 kJ,ΔG∘f(NO2)=50 kJ; R=8.3 J K−1 mol−1 and ln10=2.3 Find ΔG (in J) of the reaction at 300 K.

The correct option is D 5727χN2O4=510=12χNO2=510=12∴pN2O4=pNO2=12×20=10 atm Reaction quotient Qp=[PNO2]2PN2O4=10010=10ΔG∘reaction=2ΔG∘f(NO2)−ΔG∘f(N2O4)=2×50−100=0 We know that, ΔG=ΔG∘+RTlnQp=0+8.3×300ln10=8.3×300×2.3=5727 J