Question

When 1.045 g of CaO is added to 50.0 mL of water at 25.0 °C in a calorimeter, the temperature of the water increases to 32.3 °C. Assuming that the specific heat of the solution is 4.18 J/(g⋅°C) and that the calorimeter itself absorbs a negligible amount of heat, calculate ΔH in kilojoules/mol Ca(OH)2 for the reaction

Answer 1

1.71 kJ/mol

Step-by-step explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

`\Delta H=m* C* \Delta T`

Where,

`\Delta H` is the enthalpy change

m is the mass

C is the specific heat capacity

`\Delta T` is the temperature change

Thus, given that:-

Mass of CaO = 1.045 g

Specific heat = 4.18 J/g°C

`\Delta T=32.3-25.0\ ^0C=7.3\ ^0C`

So,

`\Delta H=1.045* 4.18* 7.3\ J=31.88713\ J`

Also, 1 J = 0.001 kJ

So,

`\Delta H=0.03189\ kJ`

Also, Molar mass of CaO = 56.0774 g/mol

`Moles=(Mass)/(Molar\ mass)=(1.045)/(56.0774)\ mol=0.01863\ mol`

Thus, Enthalpy change in kJ/mol is:-

`\Delta H=(0.03189)/(0.01863)\ kJ/mol=1.71\ kJ/mol`