00:01
Hi, in this question we need to determine enthalpy of the reaction when no reacts with no2 it gives n2o3 gaseous.
00:14
Here the enthalpy of the reaction is minus 39 .8 kj per mole.
00:20
Consider it first equation.
00:22
For the second equation, 2 moles of nitrogen dioxide gives dinitrogen tetraoxide.
00:31
Here enthalpy changes minus 57 .2 kj per mole.
00:37
This one is second.
00:40
Then 2 moles of nitrogen monoxide and oxygen gas gives nitrogen dioxide and gives enthalpy change minus 114 .2 kj per mole.
01:00
After that the fourth one reaction no gaseous plus no2 gaseous and oxygen gaseous gives n2o5 gaseous where enthalpy changes minus 166 .6 kj per mole.
01:21
Then we need to find out for dinitrogen trioxide plus dinitrogen pentoxide that gives 2 moles of dinitrogen tetraoxide.
01:36
We need to find out its enthalpy change.
01:40
Reversing equation 1 and multiplying it with 2, we get 2 moles of n2o3, 2 no gaseous plus 2 no2 gaseous.
02:02
Enthalpy change 1 is equal to 2 multiplied by 39 .8 kj per mole.
02:08
We get the value 79 .6 kj per mole...