A total of 2.00 mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 178 g of solution. The reaction caused the temperature of the solution to rise from 21.00 to 24.70 ∘C. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the solution is the same as that of pure water.

-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:

= - 2755.58 J

Finally, the enthalpy of the reaction can be determined as :

Δ =

Δ =

Δ = -1377.79 J. mol⁻¹

its first and third is your answer i did the qustion the fourth isnt right

wher is the rest of the question