For the reaction- X2O4l- 2XO2g - U - 2-1 kcal- - S-20 cal K-1 at 300KHence- - G is

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, Δ S = 20 c a l K^{- 1} a t 300 K

Hence,

Δ G

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Q. For the reaction

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

Δ U = 2.1 K c a l,

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Δ G

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Q. For the reaction,

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

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Δ S = 20 c a l K^{- 1}

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Δ G

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.
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Q. For the reaction at

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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