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Heat of rxn per mol H2O formed in acid-base reaction

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Calculate the heat of reaction per mole of water formed in the acid-base reaction. Assume that the total mass of the solution is 150g.

    75mL of 2.0M NaOH added to 75mL of 2.0M HCl in a styrofoam cup. Average initial temp = 295.65K. final temp = 308.8K

    2. Relevant equations
    ΔHrxn = qrxn = -mcΔT


    3. The attempt at a solution
    qrxn = -(141.234g)(4.186J/gK)(308.8K-295.65K) = -7.8kJ

    based on: .15 mol of HCl (5.5g) and .15 mol NaOH (6.0g) reacted to form .15 mol H2O (2.7g) and .15 mol NaCl (8.8g) therefore since the entire reaction is aqueous and the total mass of the solution is 150g: 150g -8.8g (the amount of the product that is not water) = 141.2g

    in a simpler train of thought i decided 2.70g of water were actually formed and therefore:
    qrxn = -(2.70g)(4.186J/gK)(308.8K-295.65K) = -1.49*102J

    I feel that i am grossly over thinking this, however, and I need a bit of advice
     
  2. jcsd
  3. Nov 4, 2011 #2

    Borek

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    Staff: Mentor

    Everything gets heated - so you can't ignore product and assume only water mass is important.

    Simplest approach is to assume you have 150 mL of solution before and after the reaction and assume density and specific heat capacity of water.

    That's not exactly true, but close enough. Note that density of neither solution was exactly 1 g/mL, so mass of the solution is not 150 g. But you had the same problem with your approach.
     
  4. Nov 4, 2011 #3
    Alright so "per mole of water formed" isn't really important in the calculations? Assuming the specific heat and density of the products is equal to water (I am actually instructed to do this at a later point) and assuming the mass of the solution is 150g (also instructed to do so) you're saying qrxn = -(150g)(4.186J/gK)(308.8K-295.65K) = -8.3kJ per mole of water formed?
     
  5. Nov 4, 2011 #4

    Borek

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    Staff: Mentor

    It is important, just not when you calculate amount of heat that evolved in this particular experimental setup. And you calculated this amount of heat correctly - now just check how many moles of water were produced to be able to calculate heat per mole.
     
  6. Nov 4, 2011 #5
    Oh! Okay that's a lot more clear now, thank you. So I would take the -8.3kJ and divide by moles of water (.15 seeing that HCl + NaOH → H2O + NaCl is 1:1:1:1) to get
    -8.3kJ/.15mol = -55kJ/mol
     
  7. Nov 4, 2011 #6

    Borek

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    Staff: Mentor

    I have not checked the numbers, just skimmed - and what you did looks OK now.
     
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