Question

If $S+{{O}_{2}}\to S{{O}_{2}}$ , $\Delta H$ = -298.2 kJ/mole
$S{{O}_{2}}+\dfrac{1}{2}{{O}_{2}}\to S{{O}_{3}}$ , $\Delta H$ = -98.7 kJ/mole
$S{{O}_{3}}+{{H}_{2}}O\to {{H}_{2}}S{{O}_{4}}$ , $\Delta H$ = -130.2 kJ/mole
${{H}_{2}}+\dfrac{1}{2}{{O}_{2}}\to {{H}_{2}}O$ , $\Delta H$ = -287.3 kJ/mole
The enthalpy of formation of ${{H}_{2}}S{{O}_{4}}$ at 298 K will be?
A. - 814.4 kJ/mole
B. + 814.4 kJ/mole
C. - 650.3 kJ/mole
D. - 433.7 kJ/mole

Answer 1

Hint: If we know the enthalpy of the formation of each reactant involved in the chemical reaction we can calculate the enthalpy of the product formed in the reaction. We have to do a sum of all enthalpies of the reactants involved in the chemical reaction to get the enthalpy of the formation of the product.

Complete Solution :
- In the question it is given that enthalpies of the different reactions involved in the synthesis of the sulphuric acid.
 - The chemical reaction which involves the formation of the sulphuric acid is as follows.
\[{{H}_{2}}+S+2{{O}_{2}}\to {{H}_{2}}S{{O}_{4}}\]

- Now we know the enthalpies of the formation of sulphur dioxide, sulphur trioxide, water and sulphuric acid by the reaction of sulphur trioxide with water.
- The enthalpies of the formation of sulphur dioxide, sulphur trioxide, water and sulphuric acid by the reaction of sulphur trioxide with water are -298, -98.7, -130.2 , -287.3 kJ/mole respectively.
- By adding all the above enthalpies we will get the enthalpy of the formation of sulphuric acid.
- Therefore enthalpy of the formation of the sulphuric acid is = -298 -98.7 -130.2 -287.3 = - 814 kJ/mole.
So, the correct answer is “Option A”.

Note: The sum of all the given chemical reactions in the question will give the formation of the sulphuric acid and by adding the enthalpies of the chemical reactions gives the enthalpy of the formation of the sulphuric acid.