For the reaction- X 2 O 4 l - 2 XO 2 g - U -2-1 kcal - - S -20 cal K -1 at 300 KHence- - G isA- -9-3 kcalB- 2-7 kcalC- -2-7 kcalD- 9-3 kcal

Δ H=Δ U+Δ n_g RT Given, Δ U=2.1 kcal, Δ n_g =2 R=2 × 10^ 3kcal, T=300K ∴ Δ H=2.1 + 2 × 2 × 10^ 3× 300=3.3 kcal Again, Δ G=Δ H T Δ S Given, Δ S = 20 × 10^ 3kc ...

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Byju's Answer

Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

For the react...

Question

For the reaction, $X_{2} O_{4 (l)} \to 2 X O_{2 (g)} Δ U = 2.1 k c a l, Δ S = 20 c a l K^{- 1} a t 300 K$
Hence, $Δ G$ is

2.7 kcal

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-2.7 kcal

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9.3 kcal

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-9.3 kcal

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Solution

The correct option is B -2.7 kcal
$Δ H = Δ U + Δ n_{g} R T$
$G i v e n, Δ U = 2.1 k c a l, Δ n_{g} = 2$
$R = 2 \times 10^{- 3} k c a l, T = 300 K$
$∴ Δ H = 2.1 + 2 \times 2 \times 10^{- 3} \times 300 = 3.3 k c a l$
$A g a i n, Δ G = Δ H - T Δ S$
$G i v e n, Δ S = 20 \times 10^{- 3} k c a l K^{- 1}$
On putting the values of $Δ H a n d Δ S$ in the equation, we get
$Δ G = 3.3 - 300 \times 20 \times 10^{- 3}$
$= 3.3 - 6 \times 10^{3} \times 10^{- 3} = 2.7 k c a l$

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Similar questions

Q. For the reaction

X_{2} O_{4} (l) \to 2 X O_{2} (g)

Δ U = 2.1 K c a l,

and

Δ S = 20 c a l K^{- 1}

at 300 K. Hence,

Δ G

is:

Q. For the reaction :

X_{2} O_{4} (l) \to 2 X O_{2} (g)

Δ U = 2.1 k c a l, Δ S = 20 c a l K^{- 1} a t 300 K

.
Hence

Δ G

is:

X_{2} O_{4} (l) ⟶ 2 X O_{2} (g), Δ U = 2.1 k c a l

Δ S = 20 c a l K^{- 1}

at 300 K. Then

Δ G

is:

Q. For the reaction

A_{(l)} \to 2 B_{(g)}

Δ U = 2.1 kcal, Δ S = 20 cal/K

at 300 K
Hence

Δ G

k c a l

Q. For the reaction,

X_{2} O_{4} (i) \to 2 X O_{2} (g)

Δ U = 2.1 k c a l, Δ s = 20 c a l K^{- 1}

at 300 K.

Hence,

Δ G

is:

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For the reaction, X2O4(l)→2XO2(g) ΔU=2.1 kcal, ΔS=20 cal K−1 at 300K Hence, ΔG is

For the reaction, $X_{2} O_{4 (l)} \to 2 X O_{2 (g)} Δ U = 2.1 k c a l, Δ S = 20 c a l K^{- 1} a t 300 K$
Hence, $Δ G$ is