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.

Step-by-step 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* \Delta H_f(product)]-\sum [n* \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* \Delta H_f_((HF)))+(1* \Delta H_f_((CaSO_4)))]-[(1* \Delta H_f_((CaF_2)))+(1* \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* (-271))+(1* (-1434))]-[(1* (-1220))+(1* (-814))]\\\\\Delta H_(rxn)=58kJ`

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