Question

100 ml of 2 M acetic acid was added to 100 mL of 2 M NaOH...

100 ml of 2 M acetic acid was added to 100 mL of 2 M NaOH at 23.5C and the temperature of the solution equilibrated to 36.3C. Calculate the enthalpy of neutralization for acetic acidusing the calculated heat capactiy of the coffee cup (34.87J/C)

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Answer #1

Moles of acetic acid = moles of NaOH

= 0.2 mol

Total volume of solution = 100 + 100ml = 200ml

mass of solution = 200g

increase in temperature, triangle T = 36.3 - 23.5 = 2.8degree celsius.

Heat released during the reaction = heat absorbed in temperature increase of solution and cofee cup

= mc triangle T + C calorimeter triangle T

= 2440.7J

This heat is released per 0.2 mol of the reaction.

This heat is released per 0.2 mol of the reaction.

So, heat released per 1 mol pf the reaction

12.203 KJ mol (ANSWER)

12.203 KJ mol (ANSWER)

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