CaCl2(aq)+Na3PO4(aq) -----------> Ca3(PO4)2(s) +NaCl(aq)4) If the “H” for this reaction is -45.5 kJ/mol, is the reaction endo- or exothermic?
Exothermic
Yes, the - sign tells you it is exothermic. Note that the equation you wrote should be 3NaCl.
The reaction is exothermic because the change in enthalpy (∆H) is negative (-45.5 kJ/mol). In an exothermic reaction, heat is released from the system to the surroundings.
To determine whether the reaction is endothermic or exothermic, we need to utilize the concept of enthalpy change (ΔH).
In this case, the ΔH for the reaction is given as -45.5 kJ/mol. A negative value for ΔH indicates an exothermic reaction, while a positive value indicates an endothermic reaction.
Therefore, since the ΔH is negative (-45.5 kJ/mol), we can conclude that the reaction is exothermic.