Question

3.1 mole of FeCl3 and 3.2 mol NH4SCN ARE
added to one litre of water.at equilibrium 3 mole of
of FeSCN2+ are formed. The equilibrium constant kc of the reacion Fe3+ + SCN- = FeSCN2+
will be
(1) 6.66 x 10-3
(3) 3.30
(2) 0.30
(4) 150​

Answer 1

The value of $K_c$ for the reaction is, (4) 150

Solution :  Given,

Moles of $FeCl_3$ at equilibrium = 3.1 mol

Moles of $NH_4SCN$ at equilibrium = 3.2 mol

Moles of $FeSCN^{2+}$ at equilibrium = 3 mol

Now we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

                         $Fe^{3+}+SCN^-\rightleftharpoons FeSCN^{2+}$

Initially conc.   3.1         3.2             0

At eqm.             (3.1-3)   (3.2-3)          3

                          = 0.1    =  0.2       3

The expression of $K_c$ will be,

$K_c=\frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]}$

$K_c=\frac{(3)}{(0.1)\times (0.2)}$

$K_c=150$

Thus, the value of $K_c$ for the reaction is, 150

Learn more about : Equilibrium constant

https://brainly.com/question/1266745

https://brainly.com/question/1446599

#learnwithbrainly