Answer:
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.
The actual ∆H would be greater than the calculated value of ∆H with no calorimeter.
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)
Thus, q(calorimeter) > 0(endothermic).
The actual ∆H would be greater than the calculated value of ∆H with no calorimeter.
Find more information about Enthalpy change here:
brainly.com/question/11628413
