Is breaking bonds endothermic or exothermic?
A quick way to remember this is BENDO MEX:
Breaking bonds is ENDOthermic = +ve enthalpy change
Making bonds is EXothermic = -ve enthalpy change
LS
Related Chemistry IB answers
All answers ▸Use the following information: [[[[2H2(g) + O2(g) → 2H2O(l) ∆H = −572 kJ mol−1]]]] [[[[2H2(g) + O2(g) → 2H2O(g) ∆H = −484 kJ mol−1]]]] to calculate the enthalpy change for the process: H2O(g) → H2O(l)
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