Question

For the reaction $SO_{2} +\frac{1}{2} O_{2} =SO_{3,}$ if we write $K_{p}=K_{c} \left(RT\right)^{x},$ then x becomes

• A. -1
• B. $-\frac{1}{2}$
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (B)

$-\frac{1}{2}$

Answer 2

In the relation, $K_{p}=K_{c}(R T)^{x}$,
$x=n_{p}-n_{r}$
For the reaction,
$SO _{2}(g)+\frac{1}{2} O _{2}(g) \rightleftharpoons SO _{3}(g)$
$x=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2}$