Calculating Equilibrium Constant Question

[dianegaba]

Hello,

I have a formula which is Fe(3+) + SCN (-1) --><-- Fe(SCN)(2+) and initially Fe has 0.016 mol and SCN has 0.13 mol. The volume is 500ml. At equilibrium Fe(SCN) is 0.03 mol. The temperature is 25C. And it's asking us to find the equilibrium constant.

I'm used to having different mole ratios to work from so I do not understand what to do in this question.

[Tyleralp1]

An ICE vs. Chemical table may help.

Initial mol/conc.
Consumed/Reacted amount mol/conc.
End/Final amount in mol/conc.

You should be able to deduce how much of each reacted by knowing mol ratios. Then, make sure you use final quantities of each chemical and sub in the conc values into the equation. You may need to use the gas equation to find unknown variables.

[dianegaba]

@Tyleralp1 since there all a 1:1:1 ratio I am unsure on how much I should deduce how much of each reacted

[Mellyboo]

Hello,

I have a formula which is Fe(3+) + SCN (-1) --><-- Fe(SCN)(2+) and initially Fe has 0.016 mol and SCN has 0.13 mol. The volume is 500ml. At equilibrium Fe(SCN) is 0.03 mol. The temperature is 25C. And it's asking us to find the equilibrium constant.

I'm used to having different mole ratios to work from so I do not understand what to do in this question.

R(eaction): Fe(3+) + SCN (-1) --><-- Fe(SCN)(2+)
I(itial) :        0.16mol   0.13mol.            0mol
C(hange) :  -0.03mol.    -0.03mol.       +0.03mol
E(equilibrium) : 0.13mol.   0.10mol.      0.03mol
Concentration:  0.13/0.5.  0.10/0.5.      0.03/0.5
= 0.26M Fe(3+) , 0.20M SCN(-1) and 0.06M Fe(SCN)(2+)
K = [products]/[Reactants] = 0.06/(0.20x0.26) =1.15 M^-1 (adjust for sig figs)