Question

When 0.5 gram of sulphur is burnt to a SO2 4.6 kilo joule of heat is liberated what is the enthalpy of formation of Sulphur dioxide

Answer 1

$4.6 \times 32 \div 0.5 = 294.4kj \: per \: mole$

Answer 2

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.

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

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

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

According to stoichiometry:

1 mole of sulphur gives = 1 mole of sulphur dioxide

Thus 0.016 moles of sulphur gives =$\frac{1}{1}\times 0.016=0.016$ 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 =$\frac{4.6}{0.016}\times 1=287.5kJ$

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

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