00:01
Enthalpy of formation of benzene that is 6c plus 3h2 give rise to c6h6.
00:09
Here the value of del h is equals to 48 .7 kilojoule per mole.
00:16
This is equation 1.
00:18
Then enthalpy of formation of carbon dioxide c plus o2 give rise to co2.
00:26
So the del h here is equals to minus of 393 .5 kilojoule per mole.
00:35
So this is equation 2.
00:38
Then the enthalpy of formation of h2o, h2 plus half o2 give rise to h2o.
00:49
So here the enthalpy the del h is equals to minus of 286 kilojoule per mole.
00:58
Here this is equation 3.
01:01
Then combustion of benzene that is c6h6 2c6h6 plus 15o2 give rise to 12co2 plus 6h2o.
01:25
This is equation 4.
01:29
So as seen from the equation 4 to use hess's law, we need to multiply equation 1, 2 and 3 to obtain the appropriate coefficients of c6h6, co2 and h2o so that these equations can be used to calculate the enthalpy of combustion of benzene using the hess's law.
01:49
So 12 carbon plus 6h2 give rise to 2c6h6 2c6h6 that is del h is equals to 97 .4 kilojoule...