Question

When 3 moles of the reactant A and 1 mole of the reactant B are mixed in a vessel of volume 1 L, the following reaction takes place, $A(g)+B(g)2C(g)$ . If 1.5 moles of C is formed at equilibrium, the equilibrium constant $(K_c)$ of the reaction is:

• A. 0.12
• B. 0.50
• C. 0.25
• D. 2.25
• E. 4.00

Answer 1

Answer: (E)

$A(g)+B(g)2C(g)$
a
b
0
3 mol
1 mol
0
Initially
$(a-x)$
$(b-x)$
$2x$
$\left(3-\frac{1.5}{2}\right)$
$\left(1-\frac{1.5}{2}\right)$
$=1.5$
At equilibrium
$=2.25$
$=0.25$
$2x=1.5$ $x=\frac{1.5}{2}=0.75$ $K_c=\frac{{[C]}^2}{[A][B]}$ ?..(i) Now, substituting the value of [A], [B] and [C] in the Eq. (i) $K_c=\frac{{[1.5]}^2}{[2.25][0.25]}=\frac{1.5\times 1.5}{2.25\times 0.25}$ $=4.00$