00:01
Problem 79 says to recalculate the heat of formation of each reaction from problem 69 with standard enthalpyes of formation from appendix 4.
00:12
The problem also says to compare your results from problem 69 where you used bond energies to calculate the heat of formation instead of standard enthalpyes.
00:24
So the first part of problem 69 takes hydrogen gas and chlorine gas and makes hydrochloric acid.
00:34
If you look at the appendix for in your textbook, you'll see that standard impalpies of formation for naturally occurring either elements or like these diatomic molecules that exist naturally, their standard impleps of formation are zero.
00:58
However, for a molecule like hydrochloric acid, the standard enthalpy is negative 92 kilojoules per mole.
01:11
So that means, means that the delta h for this particular reaction is negative 184 kilojoules.
01:19
And that's compared to problem 69, where bond energies were used to calculate a delta h of negative 183...
