Question

When sulphur ( in the form of $S_8$) is heated to temperature T, at equilibrium, the pressure of $S_8$ falls by 30% from 1.0 atm, because $S_8$(g) is partially converted into $S_2$(g). Find the value of $K_p$ for this reaction.

• A. $2.96$
• B. $6.14$
• C. $204.8$
• D. None of these

Answer 1

The correct option is A $2.96$

$S_8\rightleftharpoons 4S_2$

Initial pressure $1$atm $0$

At equilibrium pressure $(1-0.3)$ $4\times 0.3$

Therefore, equilibrium pressure of $S_8=(1-0.3)=0.7$ atm

And equilibrium pressure of $S_2=4\times 0.3=1.2$ atm

So, equilibrium constant $Kp=\frac{P_{S_2}^4}{P_{S_8}}=\frac{{1.2}^4}{0.7}=$ approximate $2.96$ atm${}^3$.