00:01
This question wants us to compare the delta g that we've calculated from question 62 to the ones we can obtain from the equation that relates to delta g, delta h and delta h together at 25 degrees celsius.
00:18
So we know that delta g is equal to delta h minus t delta h.
00:29
And we are looking for delta g in this case.
00:32
So therefore we calculated delta g, we calculated delta h to be equals to 64 .6 kilojoules and then minus temperature is 25 degrees celsius and now with 298 .15 kelvin multiplied by delta hs is minus 12 .7 joules per kelvin.
01:03
So we need to convert this to kilojoux because the delta h is in kilojou.
01:10
So multiply by minus 12 .7 k kilojoules divided by 1000 k.
01:20
So that should be 6 .6 .4 .6 minus 2 million 8 .5 times my name is 2 .7 ,000.
01:47
So this would be, let's see, 68 .4, 68 .4 kiverjoules.
02:04
And this is more reliable to relate delta g with temperature, as you can see the temperature time here.
02:13
The next one is to do the same team for the second calculation general problem 62...