Calculating delta H (enthalpy change) of a reaction

• Scarlet wrote on 2012-02-24 10:12
  I'm trying to find the enthalpy of a neutralization for phosphoric acid reacting with sodium hydroxide; where 25.0 ml of equimlar solutions of acid and base with an initial temperature of 22.5 degrees Celcius are mixed in an open polystyrene calorimeter. Each solution has a density of 1.00 g/mL and specific heat of 4.18 J/g degrees Celcius. The maximum temperature was 26.4 degrees Celcius.
  The above problem is a prelab question, and I'm not sure if I'm supposed to use mass of both solutions combined or just one (I think it's both, for 25 g, but I really have no clue).
  
  Also, in class, we performed an actual experiment with 50mL of .6M H3PO4 and 50mL 1.85M NaOH. (Apparently, having a limiting reactant makes experiment results more accurate). Do I have to adjust for there being less H3PO4 when finding enthalpy of neutralization of phosphoric acid? Am I supposed to include the mass of the solution in the calculation? Do I use the mass of phosphoric acid by itself (Like the moles multiplied by mass number to get atomic mass), or the product, or the product and the solution? I'm really confused at what exactly the "mass" part in the equation is supposed to be >.<
  
  Any help is appreciated.
• Lan wrote on 2012-02-24 10:39
  q=mcΔT
  
  ...urgh I would help you but my brain isn't working. Doesn't look that hard, just remember your ice table I guess ~_~
• chaolin wrote on 2012-02-24 13:26

  Quote from Scarlet;785364:

  I'm trying to find the enthalpy of a neutralization for phosphoric acid reacting with sodium hydroxide; where 25.0 ml of equimlar solutions of acid and base with an initial temperature of 22.5 degrees Celcius are mixed in an open polystyrene calorimeter. Each solution has a density of 1.00 g/mL and specific heat of 4.18 J/g degrees Celcius. The maximum temperature was 26.4 degrees Celcius.
  The above problem is a prelab question, and I'm not sure if I'm supposed to use mass of both solutions combined or just one (I think it's both, for 25 g, but I really have no clue).

  
  
  Doesn't that make both solutions water LOL? Your ΔT would then be (26.4-22.5) = 3.9 C
  
  

  Quote from Scarlet;785364:

  Also, in class, we performed an actual experiment with 50mL of .6M H3PO4 and 50mL 1.85M NaOH. (Apparently, having a limiting reactant makes experiment results more accurate).

  
  
  I don't understand what your experimental variable is? I'm assuming it's the amount of enthalpy to neutralize this stoof and make salts + water?
  
  

  Quote from Scarlet;785364:

  Do I have to adjust for there being less H3PO4 when finding enthalpy of neutralization of phosphoric acid?

  
  
  Less? Don't you have the same volume of both solutions?
  
  

  Quote from Scarlet;785364:

  Am I supposed to include the mass of the solution in the calculation?

  
  
  So your general formula for specific heat given constant pressure is what Lan said:
  
  Q = (Mass)(Specific Heat)(Change in Temperature) or Q=mCΔT
  
  Rearrange if trying to find the mass you need:
  
  m=Q/(CΔT)
  
  

  Quote from Scarlet;785364:

  Do I use the mass of phosphoric acid by itself (Like the moles multiplied by mass number to get atomic mass), or the product, or the product and the solution? I'm really confused at what exactly the "mass" part in the equation is supposed to be >.<

  
  
  It depends what Specific Heat value you use. Typically in chemistry, you would use molar heat capacity thus you want to use moles (pretty much a general rule unless you're using density I guess).
  
  ______________
  
  Maybe I'm just tired but I'm really not understanding what you're trying to do here other than calculate the total heat change of the reaction. For that, just find Q for both substances using the formula then sum them for the net change.
• Episkey wrote on 2012-02-24 20:18
  Ah. I just read your PM today - I'm not always that great when it comes to thermodynamic calculations, but I feel semi-confident in my answer. I'll post it here, that way it can be reviewed by others.
  
  I'll put my thought process below.
  
  For the pre-lab question, you do need to add up the grams of both solutions.
  Making it 50. g of solution.
  
  Also remember the equation, Heat absorbed by solution (q) = mcΔT
  
  

  
  
  • Assume density to be close to that of water, 50 mL = 50. g
    
  • Specific Heat Capacity = 4.18 J/g° C
    
  • Temperature rise (ΔT) = (26.4-22.5) = 3.9 °C
    

  
  
  All you need to do is substitute your variables in the equation to find q.
  
  Numerical answer for q:
  
  [SPOILER="Spoiler"]q = 50g x 4.18 J/g° C x 3.9°C
  = 815.1 J[/SPOILER]
  
  Now, the standard enthalpy change (∆H) is obtained by division with the amount of substance (in moles) involved.
  
  ∆H = - q/n
  
  You have q, but now you need n (number of moles).
  
  Since the number of moles of the acid and base is the same, rearrange your Molarity equation to find number of moles. (you know, M= mol/L)
  
  [SPOILER="Spoiler"]You can just assume 1M, to make the calculation easy
  n = M x L = (1.0 mol/L)(0.025 L) = 0.025 mole
  [/SPOILER]
  
  You have n and q, so .... plug them into the equation (∆H = - q/n)
  
  [SPOILER="Spoiler"]Since 815.1 J of heat is released during the neutralization of 0.025 mol, for 1 mol the amount of heat is:
  
  (- 815.1 J / .025 mole) = - 32604 J
  
  - 32604 J are given off for every 1 mole of substance. (- 32604 J/mol) [/SPOILER]
  
  Enthalpy change of neutralization given in terms of kJ/mol
  So you want your answer in kJ/mol not J/mol. Convert J to kJ.
  
  [SPOILER="Spoiler"]- 32604 J / 1000 J = - 32.604 kJ/mol
  
  Or simply, -32.6 kJ/mol (Significant figures)
  
  ΔH (neutralization) = -32.6 kJ/mol
  [/SPOILER]
  
  I hope that's right. I don't have enough time to fully answer the second part , but I hope this helps :whistle:
  
  
  
  

  Quote from chaolin;785548:

  Doesn't that make both solutions water LOL? Your ΔT would then be (26.4-22.5) = 3.9 C

  
  
  When you measure enthalpy changes for a reaction between two aqueous solutions, you can assume that the density is that of water.
  
  

  Quote from chaolin;785548:

  
  Less? Don't you have the same volume of both solutions?

  
  
  Yes, but the Molarity is different. So, the reaction would stop whenever one reactant is used up.
  
  Then H3PO4 would be the limiting reactant? I guess it depends on the chemical equation and the mole ratios >.<
• chaolin wrote on 2012-02-25 07:02

  Quote from Episkey;785651:

  When you measure enthalpy changes for a reaction between two aqueous solutions, you can assume that the density is that of water.
  
  Yes, but the Molarity is different. So, the reaction would stop whenever one reactant is used up.

  
  
  Derp derp derp derp
  
  Man, having no AP chem sucks :C
• Episkey wrote on 2012-02-25 07:27

  Quote from chaolin;786181:

  Derp derp derp derp

  
  
  It's okay lol. You wrote that post kinda late at night.
  Besides, I didn't write all that from my head. I had to look some of it up D:
  
  

  Quote from chaolin;786181:

  Man, having no AP chem sucks :C

  
  
  Best year of my life :skip:
• Maenad wrote on 2012-02-25 08:02
  I got lost at enthalpy because I don't know what that is..