Question

For the gas-phase hydration of hexafluroacetone $(C{F}_3{)}_{2}CO$:
$\left(C{F}_3{)}_{2}CO\right(g)+H_{2}O(g\left)\underset{K_b}{\stackrel{K_f}{\rightleftharpoons }}\right(C{F}_3{)}_{3}C\left(OH{)}_2\right(g)$
At ${76}^{0}C$, $K_f$ = $0.13M^{-1}{s}^{-1}$ and $K_b$ = $6.2\times {10}^{-4}{s}^{-1}$ then equilibrium constant $K_c$ is

• A. $4.77\times {10}^{-3}M$
• B. $209.7M^{-1}$
• C. $2.96\times {10}^{-7}M$
• D. $3.38\times {10}^6{M}^{-1}$

Answer 1

The correct option is A $4.77\times {10}^{-3}M$

Equilibrium constant $K_c$ is the ratio of concentration of products to that of concentration of reactants.

$K_c$ is also equal to the ratio of equilibrium constant for forward reaction, i.e. $K_f$ to that of equilibrium constant for backward reaction, i.e. $K_b$.

$\left(C{F}_3{)}_{2}CO\right(g)+H_{2}O(g\left)\underset{K_b}{\stackrel{K_f}{\rightleftharpoons }}\right(C{F}_3{)}_{3}C\left(OH{)}_2\right(g)$

$K_f=0.13Mse{c}^{-1};K_b=6.2\times {10}^{-4}se{c}^{-1}$

$K_c=\frac{K_f}{K_b}=\frac{0.13}{6.2\times {10}^{-4}}\times M$

$K_c=4.77\times {10}^{-3}M$

So, option (A) is correct.