Question

The initial temperature of the water in a constant-pressure calorimeter is 24°C. A reaction takes place in the calorimeter, and the temperature rises to 87°C. The calorimeter contains 367 g of water, which has a specific heat of 4.18 J/(g·°C). Calculate the enthalpy change (ΔH) during this reaction.

Answer 1

4739.86 J/mol.

Step-by-step explanation:

• It is known that:

The amount of heat released to water = Q = m.c.ΔT.

where, m is the mass of water (m = 367.0 g).

c is the specific heat capacity of water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 87.0°C - 24.0°C = 63.0°C).

∴ The amount of heat absorbed by released to water = Q = m.c.ΔT = (367.0 g)(4.18 J/g°C)(63.0°C) = 96645.78 J.

• Since, the reaction releases heat, so the reaction is exothermic (ΔH is negative).

ΔH = Q/n.

n = mass/molar mass = 367 g/18 g/mol = 20.39 mol.

∴ ΔH = Q/n = - (96645.78 J)/(20.39 mol) = 4739.86 J/mol.