kevinzak21 Posted September 21, 2010 Share Posted September 21, 2010 Hello! I am a high school senior taking an advanced chemistry course. I have been working on this problem for two days straight, and cannot figure out our error. Here is the problem we are given: Calculate the standard enthalpy of formation for the reaction HCL (g) + NH3 (g) ---> NH4CL (s), given the following thermochemical equations: H2 (g) + Cl2 (g) ---> 2HCL (g), standard enthalpy of formation: -184kJ (-92kJ/mol) N2 (g) + 3H2 (g) ---> 2NH3 (g), standard enthalpy of formation: -92kJ (-46kJ/mol) N2 (g) + 4H2 (g) + Cl2 (g) ---> 2NH4CL (s), standard enthalpy of formation: -628kJ (-314kJ/mol) It wants us to combine the equations using Hess's Law to arrive at the desired equation, and thus, the desired enthalpy of formation. I did so, and I arrive at the desired equation given above ( HCL (g) + NH3 (g) ---> NH4CL (s) ). However, the number I have arrived at time and time again (-176kJ/mol) is not the accepted standard enthalpy of formation for NH4CL, -314.43kJ/mol. Someone please show me the error before I bash my head against the wall any further! Thanks! Note, the book that gives the problem lists the answer I get (-176kJ/mol) as correct. I am wondering why this is not the accepted value. I'm sure it is something simple that I am missing, but I have been staring at it for much too long now and must admit I need assistance. Link to comment Share on other sites More sharing options...
Horza2002 Posted September 21, 2010 Share Posted September 21, 2010 Well Ive done the calculation and I get the same as you do (-176 kJ/mol) so I think you've got the maths right. And if thats the anwser the books giving tghe chances are that it is right. In what state was that standard enthalpy of formation given? If its not in the gaseous phase then you will have a different value to the one you have been calculating. Link to comment Share on other sites More sharing options...
dubaya311 Posted September 21, 2010 Share Posted September 21, 2010 Hess's law => delta H of the reaction = delta H of the products - delta h of the reactants. You are not calculating the enthalpy of formation of ammonium chloride, you were already given this, you are calculating the delta H of the reaction. Link to comment Share on other sites More sharing options...
mississippichem Posted September 21, 2010 Share Posted September 21, 2010 However, the number I have arrived at time and time again (-176kJ/mol) is not the accepted standard enthalpy of formation for NH4CL, -314.43kJ/mol. this number is for the solid state reaction. Your calculated value is for the aqueous solution I believe. Link to comment Share on other sites More sharing options...
Horza2002 Posted September 21, 2010 Share Posted September 21, 2010 Arr yes good point dubaya311, this is not a formation reaction, its a 'normal' reaction that you are calculating the enthalpy for. And as mississippichem said that its for the solid state, that is wht you have a different value. The value depends on the state of the reaction 1 Link to comment Share on other sites More sharing options...
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