Question

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.

Answer 1

Change in enthalpy is -51.072 kJ/mol.

Step-by-step explanation:

Molar mass of NaOH = 39.997 g/mol

So, 4.98 g of NaOH =
`(4.98)/(39.997)` moles of NaOH = 0.125 moles of NaOH

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

Heat consumed by solution =
`(m_(solution)* C_(solution)* \Delta T_(solution))` , where m is mass C is specific heat and
`\Delta T` is change in temperature.

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

= 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)

=
`-(6384J)/(0.125mol)=-51072J/mol=-51.072kJ/mol`