Chemical Equilibrium Question 6
Question 6 - 31 January - Shift 1
For reaction: $SO_2(g)+\frac{1}{2} O_2(g) \rightarrow SO_3(g)$
$K_P=2 \times 10^{12}$ at $27^{\circ} C$ and $1 atm$ pressure. The $K_c$ for the same reaction is $\quad \times 10^{13}$. (Nearest integer)
(Given $R=0.082 L atm K^{-1} mol^{-1}$ )
Show Answer
Answer: (1)
Solution:
Formula: Relation between $K_p$ and $K_c$
$SO _{2(g)}+\frac{1}{2} O _{2(g)} \rightarrow SO _{3(g)}$
$K_P=2 \times 10^{12}$ at $300 K$
$K_P=K_C \times(RT)^{\Delta n_g}$
$2 \times 10^{12}=K_c \times(0.082 \times 300)^{-1 / 2}$
$K_C=9.92 \times 10^{12}$
$K_C=0.992 \times 10^{13}$
Ans. 1
