Question

If the equilibrium constant for reaction (i) is 2, what is the equilibrium constant for reaction (ii)?

$C{O}_2\rightleftharpoons CO+\frac{1}{2}{O}_2$..... (i)
$2CO+O_2\rightleftharpoons 2C{O}_2$ .... (ii)

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. 1
• D. 2

Answer 1

The correct option is A $\frac{1}{4}$

$C{O}_2\rightleftharpoons CO+\frac{1}{2}{O}_2$ ....(i)

$K_c=\frac{\left[CO\right][O_2{]}^{1/2}}{\left[C{O}_2\right]}=2$

$2CO+O_2\rightleftharpoons 2C{O}_2$ .... (ii)

$K_c=\frac{[C{O}_2{]}^2}{\left[CO{]}^2\right[O_2]}={\left(\frac{1}{K_c}\right)}^2=\frac{1}{2^2}=\frac{1}{4}$

Hence, option A is correct.