Question

At 700 K, the equilibrium constant, $K_p$, for the reaction $2S{O}_3\left(g\right)\rightleftharpoons 2S{O}_2\left(g\right)+O_2\left(g\right)is1.8\times {10}^{-3}$ atm. The value of $K_c$ for the above reaction at the same temperature in moles per litre would be:

• A. $1.1\times {10}^{-7}$
• B. $3.1\times {10}^{-5}$
• C. $6.2\times {10}^{-7}$
• D. $9.3\times {10}^{-7}$

Answer 1

The correct option is B $3.1\times {10}^{-5}$

Given

$Kp=1.8\ast {10}^{-3}atm$

T=700K

Solution

$Kc=\frac{Kp}{(RT{)}^{\Delta n}}$

R=0.0821 lit*atm

T=700K

$\Delta n$ = No. of moles of products - No. of moles of reactants

$\Delta n$= (2+1) - 2 = 1

Putting valuse

$Kc=\frac{1.8\ast {10}^{-3}}{0.0821\ast 700}$

$=3.1\ast {10}^{-5}$

The correct option is B