How do I calculate the standard enthalpy change for the following reaction at 25 °C ?

1 answer:

4 0

MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g)

Using the standard enthalpies of formation given in the source below:

(−601.24 kJ) + (2 x −92.30 kJ) − (−641.8 kJ) − (−285.8 kJ) = +141.76 kJ
So:

MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g), ΔH = +141.76 kJ

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