Question

If 0.5gram of sulphur burnt to sulphur dioxide 4.6kilogram of heat liberated, calculate the enthalpy of formation of sulphur dioxide?

Answer 1

Answer:

The enthalpy of formation of Sulphur dioxide is -287.5 kJ/mol

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

S(g)+O_2(g)\rightarrow SO_2(g)S(g)+O

2

(g)→SO

2

(g)

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}Number of moles=

Molar mass

Given mass

\text{Number of moles of sulphur}=\frac{0.5g}{32g/mol}=0.016molesNumber of moles of sulphur=

32g/mol

0.5g

=0.016moles

According to stoichiometry:

1 mole of sulphur gives = 1 mole of sulphur dioxide

Thus 0.016 moles of sulphur gives =\frac{1}{1}\times 0.016=0.016

1

1

×0.016=0.016 mole of sulphur dioxide

Heat released when 0.016 mole of sulphur dioxide is formed = 4.6 kJ

Heat released when 1 mole of sulphur dioxide is formed =\frac{4.6}{0.016}\times 1=287.5kJ

0.016

4.6

×1=287.5kJ

As heat released is written with negative sign, enthalpy of formation of Sulphur dioxide is -287.5 kJ/mol

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