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

Answer:

Explanation:

For a chemical reaction, the enthalpy of reaction (ΔHrxn) is … ... to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g•°C). ... Both Equations 12.3.7 and 12.3.8 are under constant pressure (which ... The specific heat of water is 4.184 J/g °C (Table 12.3.1), so to heat 1 g of water by 1 ..

Answer 2

Answer : The enthalpy change during the reaction is 4.74 kJ/mole

Explanation :

First we have to calculate the heat released during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat released = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 367 g

$T_{final}$ = final temperature of water = $87^oC$

$T_{initial}$ = initial temperature of metal = $24^oC$

Now put all the given values in the above formula, we get:

$q=367g\times 4.18J/g^oC\times (87-24)^oC$

$q=96645.78J=96.6kJ$

Thus, the heat released during the reaction = 96.6 kJ

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 96.6 kJ

n = number of moles water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{367g}{18g/mol}=20.4mole$

$\Delta H=\frac{96.6kJ}{20.4mole}=4.74kJ/mole$

Therefore, the enthalpy change during the reaction is 4.74 kJ/mole