Answer:
-1377.79 J. mol⁻¹
Explanation:
Given that:
Number of moles of the compound = 2.00 mol
mass of the solution (m) = 178 g
Change in temperature (ΔT) = (24.70 - 21.00) °C
The specific heat of pure water (c)= 4.184 J/g. °C
What is the enthalpy of this reaction?
So, let's first determine the heat absorbed by the solution by using the relation:
Q = mcΔT
Q = (178 g) × ( 4.184 J/g. °C) × (24.70 - 21.00) °C
Q = 2755.58 J
The heat absorbed by the solution is equal to the heat released in the solution; As such the heat reaction can be given as:
[tex]Q_{reaction} = -Q[/tex]
= - 2755.58 J
Finally, the enthalpy of the reaction can be determined as :
Δ[tex]H_{reaction}[/tex] = [tex]\frac{Q_{reaction}}{n}[/tex]
Δ[tex]H_{reaction}[/tex] = [tex]\frac{- 2755.58 J}{ 2 mol}[/tex]
Δ[tex]H_{reaction}[/tex] = -1377.79 J. mol⁻¹
