Question

In the introduction it was stated that the following relationship holds for both an endothermic and an exothermic reaction

-(delta H)_sys = (m*(delta T)*c)_solution + ((delta T)(heat capacity calorimeter))

Based on the expected temperature chang for each reaction, derive the sign on (delta H)_sys for endothermic reaction and for an exothermic reaction with respect to the system.

Answer 1

-(delta H)_sys = (m*(delta T)*c)_solution + ((delta T)(heat capacity calorimeter))

With respect to system; solve for:

-(delta H)_sys = (m*(delta T)*c)_solution + ((delta T)(heat capacity calorimeter))

For endothermic system, expect the solution to chang ein Temperature (lower) so delta T = negative

Therefore, the delta T in HEat capacity will also be negative

So

-(delta H)_sys =negative + negative

(delta H)_sys = - (negative) = positive

Therefore, the (delta H)_sys for endothermic processes will be ALWAYS positive, i.e. it requires energy.

b)

For exothermic system, expect the solution to change in Temperature (Higher) so delta T = positive

Therefore, the delta T in Heat capacity will also be positive

So

-(delta H)_sys =positive + positive

(delta H)_sys = - (positive) = negative

Therefore, the (delta H)_sys for exothermic processes will be ALWAYS negative, i.e. it releases energy.