Question

mol-1).
2. Calculate the temperature in centigrade when vaporisation of water is in equilibrium at one atmosphers
pressure (Enthalpy of vaporisation = 40-63 10' J moll, Avap S = 108-8 J K-molº!)
H) for the transition of liquid water to steam at 100°C is 40.8 kJ mol-1 Calculate
Enthalny change (A​

Answer 1

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For vaporization of water a...

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Asked on October 15, 2019 by

Täñü Gameti

For vaporization of water at 1 atmospheric pressure, the values of ΔH and ΔS are 40.63kJmol

−1

and108.8 JK

−1

mol

−1

, respectively. The temperature when Gibb's energy change (ΔG) for this transformation will be zero, is :

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ANSWER

According to Gibb's equation, ΔG = ΔH - TΔS when ΔG = 0, ΔH = TΔS

Given, ΔH = 40.63 kJ mol

−1

= 40.63 x 10

3

J mol

−1

ΔS = 108.8 J K

−1

mol

−1

∴ T =

ΔS

ΔH

=

108.8

40.63×10

3

= 373.43K

The temperature when Gibb's energy change (G) for this transformation will be zero, is 373.4K

The zero value of the Gibb's free energy change indicates equilibrium state.