For the reaction at 298 K A g - B g- C g - D g - Ho - 29-8 kcal - - So - 0-1 kcal-KCalculate - Go and K-

The correct option is D Δ G^o = 0 ; K = 1 As we know, Δ G^o = Δ H^o TΔ S^o #160; #160; #160; #160; #160; #160; = 29.8 ( 298×0 ...

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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

For the react...

Question

For the reaction at $298 K$

$A (g) + B (g) ⇌ C (g) + D (g)$

$Δ H^{o} = 29.8 k c a l; Δ S^{o} = 0.1 k c a l / K$

Calculate $Δ G^{o}$ and $K$ .

Δ G^{o} = 0; K = 1

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Δ G^{o} = 1; K = e

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Δ G^{o} = 2; K = e^{2}

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None of these

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Solution

The correct option is D $Δ G^{o} = 0; K = 1$
As we know,

$Δ G^{o} = Δ H^{o} - T Δ S^{o}$

$= 29.8 - (298 \times 0.1)$

$= 29.8 - 29.8 = 0$

Therefore, $Δ G^{o} = 0$

The relation between $Δ G^{0}$ and $K$

$Δ G^{0}$ = $- R T l n K$

$K = 1$

So, the correct option is $A$

Suggest Corrections

Similar questions

Q. For the reaction,

A (g) + B (g) \to C (g) + D (g), Δ H^{o}

and

Δ S^{o}

are, respectively,

- 29.8 k J m o l^{- 1}

and

- 0.100 k J K^{- 1} m o 1^{- 1}

at 298 K.

The equilibrium constant for the reaction at 298 K is :

Q. Determine

Δ G^{o}

for the following reaction.

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)

;

Δ H^{o} = - 282.84 k J

[Given:

S_{C O_{2}}^{o} = 213.8

^{- 1}

mol

^{- 1}

S^{o}_{C O (g)} = 197.9

J K

^{- 1}

mol

^{- 1}

S^{o}_{O_{2}} = 205.8

J K

^{- 1}

mol

^{- 1}

]

Q. Consider the values of

Δ H^{o}

(in

k J {m o l}^{- 1}

) and for

Δ S^{o}

(in

J {m o l}^{- 1} K^{- 1}

) given for four different reactions. For which reaction will

Δ G^{o}

increase the most positive) when temperature is increased from

0^{o} C

25^{o} C

Q. In an equilibrium reaction, for which

Δ G^{o} = 0

, the equilibrium constant

K

should be equal to:

Q. The equilibrium constant for the reaction is

10

. Calculate the value of

Δ G^{o}

; (Given

R = 8.314 {J K}^{- 1} {mol}^{- 1}; T = 300 K

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For the reaction at 298KA(g)+B(g)⇌C(g)+D(g)ΔHo=29.8kcal;ΔSo=0.1kcal/KCalculate ΔGo and K.

The correct option is D ΔGo=0;K=1As we know,ΔGo=ΔHo−TΔSo =29.8−(298×0.1) =29.8−29.8=0Therefore, ΔGo=0The relation between ΔG0 and KΔG0 = −RTlnKK=1So, the correct option is A

For the reaction at $298 K$

$A (g) + B (g) ⇌ C (g) + D (g)$

$Δ H^{o} = 29.8 k c a l; Δ S^{o} = 0.1 k c a l / K$

Calculate $Δ G^{o}$ and $K$ .

The correct option is D $Δ G^{o} = 0; K = 1$
As we know,

$Δ G^{o} = Δ H^{o} - T Δ S^{o}$

$= 29.8 - (298 \times 0.1)$

$= 29.8 - 29.8 = 0$

Therefore, $Δ G^{o} = 0$

The relation between $Δ G^{0}$ and $K$

$Δ G^{0}$ = $- R T l n K$

$K = 1$

So, the correct option is $A$