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 ▸What is hybridisation?
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)
Why is there a significant difference between the radii of first and second row transition metals, where as no difference (or even a decrease) is observed between the second and third rows?
How is the bonding in NH3 and what does this entail for its electron-domain geometry and molecular geometry?