#365 The dissociation vapour pressure of NH4Cl

The dissociation vapour pressure of NH4Cl - Chemistry

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The dissociation vapour pressure of NH4Cl at 427°C is 608 kPa but at 459°C it has risen to 1115 kPa. Calculate (a) the equilibrium constant, (b) the standard reaction Gibbs energy, (e) the standard enthalpy, and (d) the standard entropy of dissociation, all at 427°C. Assume that the vapour behaves as a perfect gas and that AHⓇ and AS are independent of temperature in the range given.

II. Problems 1. At 298.15 K, Br2 is dissolved into ccls, and the mole fraction of Bra x(Bra)=0.0130. The partial pressure of Bra(g) in gas that is in equilibrium with the solution is 723.94Pa. Calculate the activity of Bra and the activity coefficient Yx under the following conditions: (1) The standard state is pure liquid Bra(l); (2) The activity coefficient of infinitely dilute Bra in Ccle solution is 1, if we have known that the vapor pressure of pure Brz(liquid) is 28397.7 Pa at 298.15 K, and the Henry constant k =54662 Pa. 2. If T = 25°C, P =100 kPa, ne mol of NaCl (B) mix with 55,5 mol of H20 (c) to form a solution, and the relationship between the V (cm') and ns can be described as follow:

3. The liquid B and liquid C can form ideal liquid mixture. At 25°C, the addition of 14 mol of pure liquid C in to the 10 mol of mixture (B+C) with X = 0.4 can form a new mixture, calculate the AG of this mixing procedure.

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SOLN 3

The balanced reaction is written as follows

Equilliribrium constant is written as

writing equillibrium pressure in terms of activity

For reference activity of pure solid is always assumed to be 1hence K reduces to:

Activity of any gas is given by

po is the standard partial pressure

Hence Total pressure will be written as :

From the balances equation we see that

Hence at 427 °C:

Hence at 459 °C:

Calculating ΔG (at 427 °C) by using the expression:

Calculating ΔH which doesn’t vary much with temperature

On Solving,

ΔH = + 161.5436788 KJ mol‐1

Now entropy can be calculated by rearranging the equation

Thus ΔS = 249.2162589 J K‐1 mol‐1

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