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 ▸Explain the effect of increasing the temperature on the rate of reaction
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)
I don't understand how to calculate initial rates of reaction based on experimental data
Which compound is a member of the same homologous series as 1-chloropropane? A. 1-chloropropene B. 1-chlorobutane C. 1-bromopropane D. 1,1-dichloropropane