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Standard XII

Chemistry

First Law of Thermodynamics

For vaporizat...

Question

For vaporization of water at $1$ atmospheric pressure, the values of $Δ H$ and $Δ S$ are $54.4 {kJ mol}^{- 1}$ and $108.8 {J K}^{- 1} {mol}^{- 1}$ , respectively. The temperature $(in Kelvin)$ when Gibb's energy change $Δ G$ for this temperature will be zero, is:

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Solution

We know,
$Δ G = Δ H - T Δ S$
If $Δ G = 0$
$Δ H = T Δ S$
Given,
$Δ H = 54.4 {kJ mol}^{- 1} = 54.4 \times 10^{3} {J mol}^{- 1} Δ S = 108.8 {J K}^{- 1} {mol}^{- 1}$
$∴ T = \frac{Δ H}{Δ S} = \frac{54.4 \times 10^{3}}{108.8} = 500 K$

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Similar questions

Q. For vaporization of water at

1

atmospheric pressure, the values of

Δ

H and

Δ

S are

40.63 k J m o l^{- 1} a n d

108.8 JK

^{- 1}

mol

^{- 1}

, respectively. The temperature when Gibb's energy change (

Δ

G) for this transformation will be zero, is :

△ H

and

△ S

for the equilibrium reaction,

H_{2} O (l) ⇌ H_{2} O (g)

at 1 atmospheric pressure are

40.63 k J m o l^{- 1}

and

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

respectively. Calculate the temperature at which the reaction occur.
Also, predict the sign of free energy for this transformation above this temperature.

The $Δ H$ and $Δ S$ for a reaction at one atmospheric pressure are +30.558 kJ and 0.066 k J $k^{- 1}$ respectively. The temperature at which the free energy change will be zero and below this temperature the nature of reaction would be:

Δ H

and

Δ S

for the system

H_{2} O (l) ⇌ H_{2} O (g)

1

atmospheric pressure are

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

and

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

respectively. Calculate the temperature at which the rates of forward and backward reactions will be the same. Predict the sign of free energy for this transformation above this temperature.

Δ H

and

Δ S

for the reaction,

{A g}_{2} O (s) ⟶ 2 A g (s) + \frac{1}{2} O_{2} (g)

are

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

and

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

respectively. Calculate the temperature at which free energy change for the reaction will be zero. Predict whether the forward reaction will be forward above or below this temperature.

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For vaporization of water at 1 atmospheric pressure, the values of ΔH and ΔS are 54.4 kJ mol−1 and 108.8 J K−1 mol−1, respectively. The temperature (in Kelvin) when Gibb's energy change ΔG for this temperature will be zero, is:

We know, ΔG=ΔH−TΔS If ΔG=0 ΔH=TΔS Given, ΔH=54.4 kJ mol−1=54.4×103 J mol−1ΔS=108.8 J K−1 mol−1 ∴T=ΔHΔS=54.4×103108.8=500 K

For vaporization of water at $1$ atmospheric pressure, the values of $Δ H$ and $Δ S$ are $54.4 {kJ mol}^{- 1}$ and $108.8 {J K}^{- 1} {mol}^{- 1}$ , respectively. The temperature $(in Kelvin)$ when Gibb's energy change $Δ G$ for this temperature will be zero, is:

We know,
$Δ G = Δ H - T Δ S$
If $Δ G = 0$
$Δ H = T Δ S$
Given,
$Δ H = 54.4 {kJ mol}^{- 1} = 54.4 \times 10^{3} {J mol}^{- 1} Δ S = 108.8 {J K}^{- 1} {mol}^{- 1}$
$∴ T = \frac{Δ H}{Δ S} = \frac{54.4 \times 10^{3}}{108.8} = 500 K$