Heat of formation question

[GRAD]

calculate Hf for C6H5OH in kcal/mol

C6H5OH +7O2----->CO2 + H2O +729.8kcal

C + O2-----> CO2 +94.4 kcal

22H2 + O2----->2H2O +136.8 kcal

Help

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[allstardentist]

is it 41.8kcal/mol?

 

[lifeisgood]

Where did you get the problem? Something doesn't seem right to me about it.

 

[GRAD]

is it 41.8kcal/mol?

yes howd you get that?

 

[lifeisgood]

Yeah how?

 

[allstardentist]

the equations seemed to be incorrectly balanced, i assumed the only correct thing was the cyclohexanol. So i just balanced it myself and solve the problem like any other

first, reverse the first equation so that the cyclohexanol is on the product side since thats what ure trying to find. so the delta H for that equation is -729.8

second, multiply the second equation by 6 to get rid of all the CO2 on the left side of the first eqn. 2(94.4)

third, multiply the third one by 3/2 to get rid of all the water on the left side of the first eqn. 136.8(3/2)

combining (adding) the 3 eqns will give you basic elements producing the cyclohexanol. also do the same with delta H. therefore the delta H for the whole process is 41.8kcal/mol.

hope it helps.

 

[GRAD]

the equations seemed to be incorrectly balanced, i assumed the only correct thing was the cyclohexanol. So i just balanced it myself and solve the problem like any other

first, reverse the first equation so that the cyclohexanol is on the product side since thats what ure trying to find. so the delta H for that equation is -729.8

second, multiply the second equation by 6 to get rid of all the CO2 on the left side of the first eqn. 2(94.4)

third, multiply the third one by 3/2 to get rid of all the water on the left side of the first eqn. 136.8(3/2)

combining (adding) the 3 eqns will give you basic elements producing the cyclohexanol. also do the same with delta H. therefore the delta H for the whole process is 41.8kcal/mol.

hope it helps.

ohhhh thanks- sorry forgot to balance, just thought it was balanced on its own....thanks bunch

 

[lifeisgood]

just figured it out too. but thanks allstar.

 

[allstardentist]

no problem