Question

9.55 g of NaOH are dissolved in 100 mL of water at constant pressure. the temperature goes from 23.6 to 47.4 degrees celcius. what is delta H for the process in KJ/mol of NaOH

Answer 1

Heat change = $mc\Delta T$ $=100\times4.2\times(47.4-23.6)=9996 J$

This is the heat produced due to addition of 9.55 gram of NaOH

moles of NaOH in 9.55 grams of NaOH =$\dfrac{9.55}{40}=0.24$

0.24 moles of NaOH produces 9996 J of energy so energy produced by 1 mole of NaOH is

$\dfrac{9996}{0.24}=41650J=41.65kJ$

Hence the $\Delta H$ for the NaOH is -41.65kJ (here sign is taken as minus because energy is released here)