Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: slu1986 on July 10, 2008, 02:15:59 PM
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When 40.0 mL of 0.200 M HCl at 21.5°C is added to 40.0 mL of 0.200 M NaOH also at 21.5°C in a coffee-cup calorimeter, the temperature of the resulting solution rises to 22.8°C. Assume that the volumes are additive, the specific heat of the solution is 4.180 J g−1 °C−1 and that the density of the solution is 1.00 g mL−1 Calculate the enthalpy change, ΔH, for the reaction:
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
The answer is -54.3 kJ, but I have no idea how to set up the equation and get that answer. If someone could please help guide me in the right direction, I would appreciate it! Thanks :)
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I'm not suprised you're having trouble, the question is criminally misleading in my opinion.
If I rewrite it (changes are in red) can you solve it now?
When 40.0 mL of 0.200 M HCl at 21.5°C is added to 40.0 mL of 0.200 M NaOH also at 21.5°C in a coffee-cup calorimeter, the temperature of the resulting solution rises to 22.8°C. Assume that the volumes are additive, the specific heat of the solution is 4.180 J g−1 °C−1 mol-1 and that the density of the solution is 1.00 g mL−1 Calculate the molar enthalpy change, ΔH, for the reaction:
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
The answer is -54.3 kJ mol-1
Better?
The answer I get for the question as it was written is -435 J, but the molar enthalpy change (which the question did not ask for) is -54.3 kJ mol-1.
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4.180 J g−1 °C−1 mol-1
You overdid here :)
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You overdid here :)
Woops! Quite right, the heat capacity was ok. Good spot Borek, 1-1 for today eh? ;)
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I'm sorry but I am still soo confused at how to set up the equation for the problem.
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How much heat evolved?
How many moles reacted?
Divide.