3 mole S of a reactant A and one mole of a reactant B are mixed in a vessel of volume 1 liter. The reaction taking place is $\mathrm{A}+\mathrm{B}\to 2\mathrm{C}$ . If 1.5 mole of C is formed at equilibrium, the value of KC is

The given equilibrium is  $\mathrm{A}+\mathrm{B}\to 2\mathrm{C}$

The ICE table for a reaction is as follows:

$\mathrm{then},\mathrm{x}=\frac{1.5}{2}=0.75$

Substitute x value in the equilibrium constant expression and then find ${\mathrm{K}}_{\mathrm{C}}$.Therefore,

${\mathrm{K}}_{\mathrm{C}}=\frac{{\left[\mathrm{C}\right]}^2}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}$

$=\frac{{\left(2\times 0.75\right)}^2}{\left(3-0.75\right)\left(1-0.75\right)}$

$=\frac{0.75\times 0.75\times 2\times 2}{2.25\times 0.25}$

=4.00

Hence, the correct answer is option “D”.