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Theloshua

  • one year ago

If 30.0 mL of 0.150 M aqueous sodium hydroxide is mixed with 30.0 mL of 0.150 M aqueous hydrochloric acid in a calorimeter at an initial temperature of 25.0 degrees Celsius, what is the enthalpy change of this reaction if the final temperature reached in the calorimeter is 27.5 degrees Celsius? NaOH + HCl yields NaCl + H2O

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  1. aaronq
    • one year ago
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    For calorimetry reactions, use the equation: \(q=m_{water}*C_p*\Delta T\) where m is the mass of water (or solution in this case) Cp is the specific heat capacity for the system (i would assume it to be the same as water here) \(\Delta T\) is change in temperature the system underwent

  2. aaronq
    • one year ago
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    oh yea and q is the heat evolved (which is equal to the change in enthalpy under constant pressure, which I assume it's valid here)

  3. Theloshua
    • one year ago
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    okayy..... so..... wait

  4. aaronq
    • one year ago
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    lol so \(C_p=4.18 ~J/~ ^oC*g\) So we have \(\sf q=m*(4.18 ~J/~ ^oC*g)*(T_{final}-T_{initial})\) we need the mass and the temperatures..

  5. Theloshua
    • one year ago
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    ummm.... but water isnt involved in the reaction...

  6. aaronq
    • one year ago
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    I know, but we're measuring the temperature of the water (solution), and that is how (indirectly) we're getting the change in enthalpy

  7. Theloshua
    • one year ago
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    but the solution isnt water XD so you dont know its heat capacity....

  8. Theloshua
    • one year ago
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    so isnt it irrelevant?

  9. aaronq
    • one year ago
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    It's common in these problems to make the simplification that the specific heat capacity is equal to that of water, unless you're told otherwise... same with the density of the solution, it's equal to that of water

  10. Theloshua
    • one year ago
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    okayy.... I hope your right... :p

  11. aaronq
    • one year ago
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    Unless you're told (in the question) otherwise, i'm right lol

  12. Theloshua
    • one year ago
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    okay

  13. Theloshua
    • one year ago
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    so ummmm hold on let me plug in the equation

  14. Theloshua
    • one year ago
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    what is the mass of the equation? :/

  15. aaronq
    • one year ago
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    the mass of the solution, the density is assumed to be the same as water's so \(\sf mass=density*volume=1~g/mL*60~mL=60~g\)

  16. Theloshua
    • one year ago
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    im confused right now....

  17. aaronq
    • one year ago
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    What are you confused about?

  18. Theloshua
    • one year ago
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    you took the density and volume of water to find the mass right?

  19. aaronq
    • one year ago
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    yep, the definition for density is: \(\sf density=\dfrac{mass}{volume}\) so i just rearranged

  20. Theloshua
    • one year ago
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    but doesnt the density and volume of water depend on the amount? or no?

  21. aaronq
    • one year ago
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    the density depends on a lot of things, most importantly temperature. Again, these problems make the simplification that the density of the solution is the same as that of water .. which is further simplified and rounded to 1 g/mL. The volume is given in the question

  22. Theloshua
    • one year ago
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    oh, you added the two volumes?

  23. aaronq
    • one year ago
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    yeah

  24. aaronq
    • one year ago
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    Thats the volume of the solution

  25. Theloshua
    • one year ago
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    okayyyy so \[Q = 60 \times 4.18 \times (27.5 - 25)\]

  26. Theloshua
    • one year ago
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    right?

  27. aaronq
    • one year ago
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    yep. theres one final thing, because this is the energy absorbed by the water it is an endothermic process, but the reaction is exothermic as energy was released. \(\sf q_{absorbed}=-q_{released}\) you need to change the sign

  28. Theloshua
    • one year ago
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    oh okay, but what about the values given (0.250 M)?

  29. Theloshua
    • one year ago
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    where do those come in?

  30. aaronq
    • one year ago
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    they don't, unless they want the molar enthalpy of the reaction. You'll come across questions that include more info than you need and you have to know what to use and what is irrelevant

  31. Theloshua
    • one year ago
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    okay, so hold on

  32. Theloshua
    • one year ago
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    The answer would be -627?

  33. aaronq
    • one year ago
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    yep, just tack on some units to that

  34. Theloshua
    • one year ago
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    okay, thnx so much!!!!

  35. aaronq
    • one year ago
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    no problem !

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