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 $