Question

Calculate the enthalpy change for the reaction: CaF2 + H2SO4 → 2HF + CaSO4 Given that enthalpy changes of formation of: AH [CaF2] = -1220 kJ mol-1. AH[H2SO4] = -814 kỤ mol-1. AH[HF] = -271 k] mol-1. AHF(CaSO4) = -1434 kJ mol-1.

Answer 1

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 $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$CaF_2+H_2SO_4\rightarrow 2HF+CaSO_4$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HF)})+(1\times \Delta H_f_{(CaSO_4)})]-[(1\times \Delta H_f_{(CaF_2)})+(1\times \Delta H_f_{(H_2SO_4)})]$

We are given:

$\Delta H_f_{(HF)}=-271kJ/mol\\\Delta H_f_{(CaSO_4)}=-1434kJ/mol\\\Delta H_f_{(CaF_2)}=-1220kJ/mol\\\Delta H_f_{(H_2SO_4)}=-814kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-271))+(1\times (-1434))]-[(1\times (-1220))+(1\times (-814))]\\\\\Delta H_{rxn}=58kJ$

Hence, the enthalpy change of the reaction is 58 kJ.