+t13 ta Great question. The heat of reaction is dependent on a reference state. In this case, the 46 kJ/mol is based "per mole of NH3". If you looked up the heat of reaction for N2 +3H2 -> 2NH3, it would be -92 kJ/mol, but since that is to produce 2 moles of NH3, you would divide the number by 2, therefore getting the same answer. Does that make sense?
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Thank you very much only i love thermodynamic from this channel thanks alot for your effort
why do not we multiply the heat of reaction for amonia at 25C by 2 since 2 moles of NH3 are formed ?
+t13 ta Great question. The heat of reaction is dependent on a reference state. In this case, the 46 kJ/mol is based "per mole of NH3". If you looked up the heat of reaction for N2 +3H2 -> 2NH3, it would be -92 kJ/mol, but since that is to produce 2 moles of NH3, you would divide the number by 2, therefore getting the same answer. Does that make sense?
LearnChemE @ 6:00 cool down the "reactant" stream
If we would Not remove the heat, what would be the temperature increase assuming an adiabatic process? Is it T_end= (q/(m*c_p)) + 350?
Thanks
this is a bad explanation , i m doing the replacement of the variables and i cant get the final result dont know why
the heat of reaction calculation , we must multiply heat of formation of ammonia by stoic. coeff. of ammonia which was not done here !!!!!!!