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cytokinase

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A scientist adds potassium hydroxide to nitric acid, which yields the following reaction:

HNO3 + KOH --> KNO3 + H2O

Data:
Mass KOH = 4.6 g
Volume of nitric acid solution = 250 mL
T1 = 21 degrees-C
T2 = 29.3 degrees-C

Calculate the MOLAR ENTHALPY of neutralization of KOH

10 for best..
 
First you have to convert to moles. For KOH, appx 4.6/57 moles or 0.081 moles. Assuming the volume of solution is unchanged, its molarity is 0.081 moles/0.25 L = 0.323M.

The heat genereated is m * Cp * delta T. m = 255 g, Cp is assumed that of water, 4.18 joule/gram -deg C. delta T = 29.3-21. So heat generated= 8850 J or 8.85 kJ. Putting this on a per-mole basis, the heat generated, which is appx the molar enthalpy of neutralization (ignoring minor PV adjustments) is 8.85 kJ/0.323 M = -27.4 kJ/mole. I believe the negative sign is used when heat is generated.
 
First you have to convert to moles. For KOH, appx 4.6/57 moles or 0.081 moles. Assuming the volume of solution is unchanged, its molarity is 0.081 moles/0.25 L = 0.323M.

The heat genereated is m * Cp * delta T. m = 255 g, Cp is assumed that of water, 4.18 joule/gram -deg C. delta T = 29.3-21. So heat generated= 8850 J or 8.85 kJ. Putting this on a per-mole basis, the heat generated, which is appx the molar enthalpy of neutralization (ignoring minor PV adjustments) is 8.85 kJ/0.323 M = -27.4 kJ/mole. I believe the negative sign is used when heat is generated.
 
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