00:01
So for this question, we're told to consider the reaction of 6 no2 gas plus 8 nh3 gas to yield s7n2 plus 12 h2o gas.
00:13
And we're asked to determine the standard heat of the reaction for this particular reaction using the standard enthalpies of formation.
00:28
So the first thing you have to do is actually look up the standard heats of formation for each of the compounds.
00:38
So the standard heat of formation for no2 gas is positive 33 .2 kj per mole.
00:47
The standard heat of formation for ammonia gas is negative 45 .9 kj per mole.
00:56
Nitrogen is in its elemental form, right? nitrogen exists as a gas under those standard conditions.
01:06
So it has a zero for its heat of formation because the heat of formation for any element is zero.
01:14
And then for water as a gas, the heat of formation was negative 241 .8 kj per mole.
01:22
So now what we're going to utilize is that the heat of the reaction is going to be equal to the sum of the number of moles times the heat of formation of the product minus the sum of the number of moles times the heat of formation of the reactants.
01:46
So it's going to be the products minus the reactants.
01:50
So what we're going to do then is we're going to put together, so we're going to do products first.
01:59
So we're going to do 7 times 0, which is the heat of formation of nitrogen, plus 12 times negative 241 .8 kj per mole.
02:20
That sum minus our reactants, which is then going to be 6 for nitrogen dioxide times positive 33 .2 kj per mole.
02:38
And that sign is very important...