Question

Calculate the enthalpy change for the reaction: CaF2+H2SO42HF+ CaSO4 Given that enthalpy changes of formation of: AHi[CaF2] = -1220 kJ mol1. AHf[H2SO4] = -814 kJ mol1. AHi[HF] = -271 kJ mol1. AHi[CASO4] = -1434 kJ mol1

Answer 1

Enthalpy change of the reaction = 58 kJ/mol

Step-by-step explanation:

Enthalpy of a reaction is calculated as:

`\Delta H = \Delta H_(Product) - \Delta H_(Reaction)`

`\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)\\`

For the given reaction,

`CaF_(2) +H_2SO_4 \rightarrow 2HF + CaSO_4`

`\Delta H = (2* \Delta H_(HF) + \Delta H_(CaSO_4)) - (\Delta H_(CaF_2) + \Delta H_(H_2SO_4))`

`\Delta H=(2* -271 -1434) - (-1220 - 814))\;kJ/mol`

= (-1976 + 2034) kJ/mol

= 58 kJ/mol