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Enthalpy Of Formation

  1. Sep 9, 2012 #1
    1. The problem statement, all variables and given/known data
    We already know that standard heat or enthalpy of formation of a substance is the heat change in forming one mole of the compound from its element in their standard states.
    Thus in the reaction the enthalpy of formation 2 mole of NH3 is 92kJ. Therefore the standard DH of formation = 92kJ/2= 46KJmol-1. What am now asking is this: why should the standard enthalpy of formation of ammonia be positive? Because I know that standard enthalpy of NH3 is suppose to be negative not positive.
    Given that for the reaction,
    N2(g)+3H2(g)---->
    <----2NH3(g) deltaH=92kJ. What is the enthalpy of formation of ammonia from its elements?


    2. Relevant equations



    3. The attempt at a solution

    We already know that standard heat or enthalpy of formation of a substance is the heat change in forming one mole of the compound from its element in their standard states.
    Thus in the reaction the enthalpy of formation 2 mole of NH3 is 92kJ. Therefore the standard DH of formation = 92kJ/2= 46KJmol-1. What am now asking is this: why should the standard enthalpy of formation of ammonia be positive? Because I know that standard enthalpy of NH3 is suppose to be negative not positive.
     
    Last edited: Sep 9, 2012
  2. jcsd
  3. Sep 9, 2012 #2

    AGNuke

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    So you are having problem with sign convention?

    Well, it is negative sign. You might have been given the absolute value. Sign just shows whether the reaction would be is endothermic or exothermic.
     
  4. Sep 9, 2012 #3
    But atleast, the sign for enthalpy of formatation of 2 mole of NH3 is suppose to be -92, so that when we get the standard enthalpy of formation for just 1 mole of NH3, it will become -46kJ/mol instead of +46KJ/mol. This is because the reaction leading to the formation of NH3 is exothermic. There must have been an error (typo) by the authors of problem as they wrote +92kJ as the enthalpy of formation of 2 mole of NH3 instead of -92kJ. Do you agree with me on that?
     
  5. Sep 10, 2012 #4

    AGNuke

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    The authors might have given the absolute value of enthalpy of formation. So, it might be the case.
     
  6. Sep 10, 2012 #5
    So then what is the meaning of the word "absolute value" as it relates to this concept?
     
  7. Sep 10, 2012 #6

    AGNuke

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    Value without sign convention used.
     
  8. Sep 10, 2012 #7
    I think Is important at this juncture that I post the choices:
    A. -46kJmol-1 B. -184kJmol-1 C. -92kJmol-1 D. 184kJmol-1 E. +46kJmol-1
    If you were in my shoe, what option will you choose?
     
  9. Sep 10, 2012 #8
    If these are the exact terms that the question was expressed in, then you have no choice but to put down option (E.) even though you know it to be wrong/unrealistic. The teacher cannot mark this (wrong) answer wrong. If the test paper is not strict multiple choice, then you should qualify your answer by briefly writing your concerns about a misprint in the question, because you are quite right.

    I am a retired university teacher. As a student I always found multiple choice tests very unfair for just this sort of reason. Often an answer to a question must be qualified, not only for a misprint. For example, if no temperature is given, am I supposed to assume 0°C, 25°C, or 0 K? Or if, for whatever reason, none of the answers offered is close to the one I get and I think I can see where an unjustified approximation has been used. As a teacher I tried to avoid them.

    Dealing with multiple choice papers is often more a matter of game tactics than science. My advice to you in terms of game tactics is to choose the answer (E.) and to discuss your problem with it if the paper allows for that possibility.
     
  10. Sep 10, 2012 #9
    You know what? This question is actually a worked question with answers. I was actually trying my hands on it to test my knowledge. The funny thing about the question, is that the answer that was provided for the question is choice A. -46kJ/mol which the correct answer. But if you work strictly with the data given, you are going to have E. +46KJ/mol instead of -46kJ/mol.
     
  11. Sep 10, 2012 #10

    AGNuke

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    Most probably a typing error in the question on their part, nothing to beat the bush around.
     
  12. Sep 10, 2012 #11
    Thanks for helping!
     
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