00:01
Hello, my name is john, and for this question, we have to find the k -equilibrium constant for the formation of f -e -s -c -n -2 -plus.
00:10
So since this is the formation of this following compound, i have the following equation in and out, and it's already balanced.
00:18
So let's go ahead and start doing simple ice table procedures.
00:22
So we're given the molarity initial concentrations of both reactants, but we're given the equilibrium concentration of the concentrations of the reactants.
00:31
Compound of the product.
00:34
So we have 5 .72 times 10 to the negative fourth malarity.
00:43
But then we have 2 .86 times 10 to the negative fourth malarity.
00:48
And that's in equilibrium with some unknown number.
00:54
So we have if we followed through simple ice table procedures, we have minus x, minus x and plus x.
01:03
Because of because of the fact that the co -ofeiting.
01:13
Are all one so these all become only minus x and plus x so for this since we know that the concentration of f e scn 2 plus at equilibrium is 4 .46 times 10 to negative fifth malarity and we know that the value of the equilibrium of f e sc and 2 plus is only x as shown from the ice table we can say that x is equal to 4 .46 times 10 to the negative fifth as well...