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)
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)
Get Answers For Free
Most questions answered within 1 hours.