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

Answer:

1.71 kJ/mol

Explanation:

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

$\Delta H=m\times C\times \Delta T$

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,  

$\Delta H=1.045\times 4.18\times 7.3\ J=31.88713\ J$

Also, 1 J = 0.001 kJ

So,  

$\Delta H=0.03189\ kJ$

Also, Molar mass of CaO = 56.0774 g/mol

$Moles=\frac{Mass}{Molar\ mass}=\frac{1.045}{56.0774}\ mol=0.01863\ mol$

Thus, Enthalpy change in kJ/mol is:-

$\Delta H=\frac{0.03189}{0.01863}\ kJ/mol=1.71\ kJ/mol$

Answer 2

The enthalpy change for the given reaction is 1.71 kJ/mol. The enthalpy is the sum total of the system's internal energy and the product of its pressure and volume.

How to calculate Enthalpy in kJ/ mol?

To calculate the enthalpy change in kJ/mol first, calculate the moles of CaO,

$n = \rm \dfrac { mass}{molar mass}\\\\n = \rm \dfrac { 1.045 \ g}{56.07 \ g/mol}\\\\n = 0.018 mol.$

The change in enthalpy can be calculated by the formula,

$\Delta H = mc\Delta T$

Where,

$m$ - mass  = 1.045 g

$c$ - specific heat =  4.18 J/g°C

$\Delta T$ -  temperature change  = 32.3-25 = 7.3°C

Put the values in the formula, we get,

$\Delta H = 0.031 \rm \ kJ$

Thus enthalpy change in kJ/mol,

$\Delta H ={\rm \dfrac {0.031 \ kJ}{ 0.018 \ mol}}\\\\\Delta H = 1.71 \rm \ kJ/mol$

Therefore, the enthalpy change for the given reaction is 1.71 kJ/mol.

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