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Heat of reaction

  1. May 26, 2005 #1
    Hello again folks.
    Im working on my chemistry work, i just completed an experiment.. Heat of Reaction. Now I am working on some questions for it... now I am stuck
    They are asking me, The amount of NaOH moles.... Im not sure how to calculate this... throughout the experiment i was mixing 5.5g of NaOH to 200mL of water and to begin with I made a solution 4.00ml 1.0mol/L NaOH solution. Hmm well i have three tables to fill out for 3 different reactions Ill show just the results for the first one, and if someone could help me with this one im sure i could figure out the other two.

    Initial temperature of water: 25 C
    Final Temperature of water: 30 C
    Mass of NaOH: 5.5g
    Volume of water 200mL
    Temperature change: 6 C
    Mass of solution: 200g ( since i had 1.0mol/L solutions)
    Heat change: 5.016kJ (used equation deltaH=m * delta T* Q (which i was given to be 4.18 x 10^-3 kJ/g C

    Now i need the amount of NaOH(moles)
    and Average H (kj/mol NaOH)
  2. jcsd
  3. May 26, 2005 #2


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    If you're asking for the enthalpy change/mole of sodium hydroxide, simply divide the heat quantity by the number of moles of sodium hydroxide, that will be your answer, in most cases however they want the change in enthalpy per mole of reaction. I'm assuming that you know how to calculate the moles of sodium hydroxide, if you don't...well that would be sad.
  4. May 26, 2005 #3
    Last edited: May 26, 2005
  5. May 26, 2005 #4


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    The molar mass of sodium hydroxide is approximately forty, otherwise you were right in calculating the moles of the compound. You do see that by varying the amount of sodium hydroxide the heat of reaction will differ, that's why it is important to factor in the moles of compound or reaction.

    It seems that from your original post that the heat of reaction is 5.016 kJ, be sure to consider the sign. If your teacher actually wants the answer in terms of moles of sodium hydroxide then you can simply divide this value by the moles of sodium hydroxide (you seem to have the right idea here
    I don't know where the 4.59 came from. Remember to adjust the moles of sodium hydroxide value.
  6. May 26, 2005 #5

    I see now.. thanks soo much :smile: (the change in #'s i realized I had the wrong answer thats all...)
  7. May 27, 2005 #6


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    good, glad you figured it out.
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