The change in enthalpy [∆ {H}] in {kJ} {mol}^{-1} for the reaction is {Mg}+2 {F} \to {Mg} {F}_{2} Given: EA of {F}=328 {kJ} {mol}{ }^{-1}, {IE}_{1} of {Mg}=737 {kJ} {mol}^{-1}, {IE}_{2} of {Mg}=1451 {kJ} {mol}^{-1}

TS EAMCET 18-July-2022 Shift 2 Solved Paper

Section: Chemistry

Question : 124 of 160

Marks: +1, -0

The change in enthalpy [∆H] in kJmol−1 for the reaction is Mg+2F→MgF2
Given: EA of F=328kJmol‌−1,IE1 of Mg=737kJmol−1,IE2 of Mg=1451kJmol−1

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