00:01
So first, we have to calculate the delta -h for the reaction.
00:05
So as we know, how do we get delta -h? we'll use the delta -h of product side minus delta -h of reactant side.
00:11
So we have two, so two times the delta -h of k -o -h was negative 481, plus the delta -h of h, which is 0, minus the left side, 2 times the k, which is 0, plus 2 times the delta -h -h -2 -0, which is negative 286.
00:28
If you do that, you'll end up with negative -8 -8 -6.
00:34
390 kilojoules per mole for the reaction for the delta h of the reaction now we have the energy so here is the heat released per one more potassium dissolved in water but we have 5 grams of k is dissolved in water so how many moles is that for 5 grams so 5 divided by the molar mass of about 40 grams per mole, we'll eventually get 0 .13 mole.
01:05
And if, so for two moles of k releases 390 kilograms of heat, for 1 .27, 2 .13 modes of k releases x kilojoules of p.
01:18
So we'll use this, this times this, divided by the 2mo.
01:26
It's just a basic conversion, and we will get 25 kilojoules...