When 4.98 g of NaOH was dissolved in 52.79 g of water in a calorimeter at 23.7 oC, the temperature of the solution went up to 50.1 oC. What is the enthalpy change in kJ/mole of sodium hydroxide? Assume the specific heat of the mixture is the same as water.

Respuesta :

Answer:

Change in enthalpy is -51.072 kJ/mol.

Explanation:

Molar mass of NaOH = 39.997 g/mol

So, 4.98 g of NaOH = [tex]\frac{4.98}{39.997}[/tex] moles of NaOH = 0.125 moles of NaOH

Total mass of solution = (4.98+52.79) g = 57.77 g

Heat consumed by solution = [tex](m_{solution}\times C_{solution}\times \Delta T_{solution})[/tex] , where m is mass C is specific heat and [tex]\Delta T[/tex] is change in temperature.

So, heat consumed by solution = [tex][57.77g\times 4.186\frac{J}{g.^{0}\textrm{C}}\times (50.1-23.7)^{0}\textrm{C}][/tex]

                                                    = 6384 J

It is an exothermic process as temperature increases during dissolution. Hence change in enthalpy should be negative.

Change in enthalpy = -(heat consumed by solution)/(no. of moles of NaOH)

                                 = [tex]-\frac{6384J}{0.125mol}=-51072J/mol=-51.072kJ/mol[/tex]