Respuesta :
Answer: The enthalpy change of the reaction is 58 kJ.
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
For the given chemical reaction:
[tex]CaF_2+H_2SO_4\rightarrow 2HF+CaSO_4[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H_{rxn}=[(2\times \Delta H_f_{(HF)})+(1\times \Delta H_f_{(CaSO_4)})]-[(1\times \Delta H_f_{(CaF_2)})+(1\times \Delta H_f_{(H_2SO_4)})][/tex]
We are given:
[tex]\Delta H_f_{(HF)}=-271kJ/mol\\\Delta H_f_{(CaSO_4)}=-1434kJ/mol\\\Delta H_f_{(CaF_2)}=-1220kJ/mol\\\Delta H_f_{(H_2SO_4)}=-814kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H_{rxn}=[(2\times (-271))+(1\times (-1434))]-[(1\times (-1220))+(1\times (-814))]\\\\\Delta H_{rxn}=58kJ[/tex]
Hence, the enthalpy change of the reaction is 58 kJ.
