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

Answer:

Enthalpy change of the reaction = 58 kJ/mol

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,

$CaF_{2} +H_2SO_4 \rightarrow 2HF + CaSO_4$

$\Delta H = (2\times \Delta H_{HF} + \Delta H_{CaSO_4}) - (\Delta H_{CaF_2} + \Delta H_{H_2SO_4})$

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

                   = (-1976 + 2034) kJ/mol

                    = 58 kJ/mol