00:01
Hello everyone, let's start the question.
00:03
In this question, we are given with the equilibrium reaction such that we have the value of equilibrium constant for this as 699.
00:12
We are asked to calculate the ratio of i3 negative concentration divided by the concentration of iodine.
00:27
So, we have the value given for the concentration of iodine.
00:32
Such that we have 2 multiplied by 10 raised to the power negative 2 moles per it and the volume is 1 litre.
00:42
So we will get the moles as 2 multiplied by 10 raised to the par negative 2 molar and the concentration of potassium iodide is also the same as the concentration of iodide ion which is 2 multiplied by 10 raised to the power negative 1 molar.
00:59
So now if we proceed further we have the initial concentration this is for iodine, ion then iodine gas this will also be an aqua state here and it is leading to the formation of i3 negative so now the initial concentration initial concentration is 2 multiplied by 10 to the power negative 1 molar.
01:40
For this, for iodine it is 2 multiplied by 10 raised to the 5 negative 2 molar and it is 0 for i 3 negative.
01:48
Then if we write the concentration at equilibrium including the change it will become 0 .2 from this we will subtract x .02 from this we will subtract x and x will be for i3 negative.
02:07
So now we can write the expression...