For the reaction: PCl5(g) = PCl3(g) + Cl2(g) We initially have 0.1 bar of PCl5,

ID: 2104114 • Letter: F

Question

For the reaction:

PCl5(g) = PCl3(g) + Cl2(g)

We initially have 0.1 bar of PCl5, 0.1 bar of PCl3 and 0.1 bar of Cl2 (note p =1 bar).

What are (i) the direction of spontaneous reaction, (ii) the equilibrium
constant, and (iii) the equilibrium partial pressures at 298 K?

Hint for (iii): You can save some time here by assuming that the equilibrium
constant is so small that it may be considered zero.

You will need these data:

Delta fG    (PCl5(g)) = -305.0 kJ mol-1
Delta fG    (PCl3(g)) = -267.8 kJ mol-1
Delta fG    (Cl2(g)) = 0 kJ mol-1

1)Delta fG of reaction = -267.8 + 305 = 37.2 kJ mol-1

this positive, hence the reaction is spontaneous in the backward direction

2)DGo = - R T ln K

lnK = -37200/ (298*8.314) =-15.015

K=e^-15.015 =3.013 *10^-7

3)as said in the question , let us consider the euilibrium constant to be zero

0=[ PCl3-x ] *[Cl2-x] / [PCl5+x]

so as the euilibrium constant is zero ,the concentration of PCl3 and Cl2 would be zero .

Thus [PCl5] =0.1 +0.1 = 0.2 bar

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