Question

The equilibrium mixture for ${2SO}_2+O_2\rightleftharpoons {2SO}_3$ present in I L vessel at 600C contains 0.50, 0.12, and 5.0 moles of $S{O}_2,O_2$ and $S{O}_3$ respectively.
A) Calculate $K_C$ for the given change at ${600}^{0}C$.
B) Also calculate $K_P$.

• A. A) $K_C=33.33$
  B) $K_P=11.62{\left(atm\right)}^{-1}$
• B. A) $K_C=333.33$
  B) $K_P=7.5{\left(atm\right)}^{-1}$
• C. A) $K_C=213.22$
  B) $K_P=1.26{\left(atm\right)}^{-1}$
• D. A) $K_C=833.33$
  B) $K_P=11.62{\left(atm\right)}^{-1}$

Answer 1

The correct option is D A) $K_C=833.33$

B) $K_P=11.62{\left(atm\right)}^{-1}$
${2SO}_2+O_2\rightleftharpoons {2SO}_3$
0.5 0.12 5.0 (Moles at equilibrium)
Volume = 1 L
A) $K_C=\frac{{{[SO}_3]}^2}{{{[SO}_2]}^2\left[O_2\right]}=\frac{{\left(5\right)}^2}{{\left(0.5\right)}^2\left(0.12\right)}$
$K_C=833.33$
B) $K_P=K_C{\left(RT\right)}^{△n}=833.33\times (0.0821\times 873{)}^{-1}=11.62{\left(atm\right)}^{-1}$