A student places 25.0 mL of a .300 M NaOH at 25.0C in a coffee-cup calorimeter.

Question

A student places 25.0 mL of a .300 M NaOH at 25.0C in a coffee-cup calorimeter. To the same calorimeter, the student adds 25.0 mL of a 0.300 M HCl at 25.0C and stirs well. After stirring the final temperature of the misture is 27.9C. Assume the total volume is the sum of the individual volumes and that the final solution has the same density as water and the same specific heat as water.

A) Determine the q value for this reaction in kJ. Answer: 0.61 kJ

B) Detemrine the the HRxn in kJ/mol. Answer: -81 kJ/mol

I know how to do part A but I keep getting a different answer for part B.

Explanation / Answer

part B)

it is very easy to calculate HRxn once we get q value

moles of NaOH = moles of HCl = 25 x 0.3 / 1000 = 7.5 x 10^-3

HRxn = - q / n

           = - 0.61 / 7.5 x 10^-3

           = -81 kJ / mol

HRxn = - 81 kJ / mol

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