Ni+2 + SCN- <--> Ni(SCN)+
A reaction flask initially contains .0006 Ni+2 and .00002 M SCN-. AFter equilibrium is reached, it is found that the concentration of Ni(SCN)+ is 6.0x10^-6 M. Complete an ICE box table.
To complete an ICE (Initial, Change, Equilibrium) box table for the given reaction, we need to analyze the stoichiometry of the reaction and determine the changes in concentrations of reactants and products when equilibrium is reached.
The balanced equation for the reaction is as follows:
Ni+2 + SCN- ⇌ Ni(SCN)+
The initial concentrations of the reactants are given:
[Ni+2] = 0.0006 M (initial concentration of Ni+2)
[SCN-] = 0.00002 M (initial concentration of SCN-)
We are also given the concentration of the product at equilibrium:
[Ni(SCN)+] = 6.0x10^-6 M (concentration of Ni(SCN)+ at equilibrium)
To determine the changes in concentrations, we can use the stoichiometry of the balanced equation:
Ni+2 + SCN- ⇌ Ni(SCN)+
Initial: 0.0006 0.00002 0
Change: -x -x +x
Equilibrium: 0.0006 - x 0.00002 - x x
As the reaction proceeds, the concentration of Ni+2 and SCN- decreases by x, while the concentration of Ni(SCN)+ increases by x.
In the equilibrium expression, Kc, the balanced equation coefficients become exponents. Since the expression for Kc is not given in the question, we cannot calculate the value of Kc or x without further information.