At a certain temperature, the Kp for the decomposition of H2S is 0.714. H2S(g) <-----> H2(g)...

At a certain temperature, the Kp for the decomposition of H2S is 0.714.

H2S(g) <-----> H2(g) + S(g)

Initially, only H2S is present at a pressure of 0.102 atm in a closed container. What is the total pressure in the container at equilibrium?

Expert Solution

H2S(g) <-----> H2(g) + S(g)

initially 0.102 atm 0 0

change x x x

equil 0.102-x x x

Kp = x^2/(0.102-x) = 0.714

x = 0.0905

total pressure at equilibrium = 0.102-x+x+x

= 0.102+x

= 0.102+0.0905

= 0.1925 atm

queen_honey_blossom answered 2 years ago

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