Chemical Equilibrium

JEE Main 2015 Questions › Physical Chemistry questions › Chemical Equilibrium

Two solids components A and B dissociate into gaseous products at 20Â°C as
i) A(s) â†’ A'(g) + H₂S(g)
ii) B(s) â†’ B'(g) + H₂S(g)
At 20Â°C pressure over the excess solid A is 50 mm Hg and that over excess solid B is 68 mm Hg.
find
a) The dissociation constants of A and B
b) the ratio of the number of moles of A' and B' in the vapor phase over a mixture of solids A and B
c) The total pressure of gas over the solid mixture.

#Chemistry #Physical Chemistry

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Vivek @ Born this Way ·2012-04-04 01:51:38

The dissociation constant is the Equilibrium constant I presume. Now

a)
K_p = P2 . P2 = P²4 , Where P is the total pressure as given and since the molefraction of each component is 1/2.

This expression should be written for both A and B, therefrom you'll get K_p_A and K_p_B

When both are mixed,

A(s) â†’ A'(g) + H₂S(g)
x x+y

B(s) â†’ B'(g) + H₂S(g)
y x+y

Again write K_P_A = P_T x2x+y . P_T(x+y)2x+y and similarly K_p_B , Now you know these quantities and hence you can solve them for x and y (since there are two relations.)

And Subsequently you'll get P_T = Total Pressure.

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