Question

4). One mole of monoclinic sulfur at 25C was placed in a constant-pressure calorimeter whose heat capacity (C) was 1620 J/K. The temperature of the calorimeter increased by 0.150 Co when the sulfur changed from the monoclinic to the orthorhombic form. Calculate the enthalpy change for the process S(monoclinic)  S(orthorhombic).

Answer 1

Answer: The enthalpy change of the reaction is -243 J/mol

Explanation:

The heat released by the reaction is absorbed by the calorimeter and the solution.

The chemical equation used to calculate the heat released follows:

$q=c\times \Delta T$

where,

c = heat capacity of calorimeter = 1620 J/K

$\Delta T$ = change in temperature = $0.150^oC=0.150K$   (Change remains same)

Putting values in above equation, we get:

$q=1620J/K\times 0.15K=243J$

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

$S\text{ (monoclinic)}\rightarrow S\text{ (orthorhombic)}$

We are given:

Moles of monoclinic sulfur = 1 mole

• To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,  

q = amount of heat released = -243 J

n = number of moles = 1 mole

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-243J}{1mol}=-243J/mol$

Hence, the enthalpy change of the reaction is -243 J/mol