Answer:
Enthalpy change of the reaction = 58 kJ/mol
Explanation:
Enthalpy of a reaction is calculated as:
[tex]\Delta H = \Delta H_{Product} - \Delta H_{Reaction}[/tex]
[tex]\Delta H_{CaF_2} = -1220\;kJ mol^{-1}\\\Delta H_{H_2SO_4} = -814\;kJ mol^{-1}\\\Delta H_{HF}=-271\;kJ mol^{-1}\\\Delta H_{CaSO_4} = -1434\;kJ mol^{-1}\\[/tex]
For the given reaction,
[tex]CaF_{2} +H_2SO_4 \rightarrow 2HF + CaSO_4[/tex]
[tex]\Delta H = (2\times \Delta H_{HF} + \Delta H_{CaSO_4}) - (\Delta H_{CaF_2} + \Delta H_{H_2SO_4})[/tex]
[tex]\Delta H=(2\times -271 -1434) - (-1220 - 814))\;kJ/mol[/tex]
= (-1976 + 2034) kJ/mol
= 58 kJ/mol
