The answers are:
Exothermic changes: the system heats the surrounding. the heat absorbed by
the surroundings increase the freedom of motion of the particles which
increases the entropy of the surrounding. therefore qsys < 0 , qsurr > 0
and delt S > 0 .
the explanation:
the exothermic reaction is a chemical reaction which releases heat, and the
heat is one of its products in the reaction. So the system here released heat
and the surroundings will absorb it.
-in the exothermic reaction, the enthalpy in the surroundings will increase,
as ΔSsurr = -ΔH/T and when ΔH in the exothermic reaction has a negative
sign( - ) so, ΔSsurr has to be +Ve ( positive sign) ∴ ΔSsurr > 0 .
-and when the system released the heat and the surroundings absorbed it this affects the freedom of motion of the particles and make it increase and this will affect the entropy and make it increase.
- in exothermic reaction when the system released heat and the surroundings absorbed it will increase the thermal disorder in the surrounding which makes
ΔSsurr >0 and qsys <0 & qsurr >0 as qsurr = - qsys
when ΔSsurr is proportional to the amount of heat transferred
∴ΔSsurr α qsurr and ΔSurr α - q sys
so when ΔSsurr has a positive sign in the exothermic reaction so qsurr >0
and qsys<0.
