If a student performs an exothermic reaction in a calorimeter, how does the calculated value of ΔH (Hcalc) differ from the actual value (Hactual) if the heat exchanged with the calorimeter is not taken into account?

The actual ∆H would be greater than the calculated value of ∆H with no calorimeter.

Explanation:

The amount of heat changed during this process at a fixed pressure is termed Enthalpy

enthalpy change ∆H = ∆E + P∆V

∆E = internal energy change

P = fixed pressure

∆V = change in volume

When energy is absorbed during reaction, it is called endothermic reaction.

Endothermic reaction carried out in the calorimeter and enthalpy change for the reaction. Since we have that

q(surrounding) = q(solution)+q(calorimeter)

Therefore, q(calorimeter) > 0(endothermic).

The actual ∆H would be greater than the calculated value of ∆H with no calorimeter.

i think its 5049.

i hope this !