Question

Consider the reactions:
i).${\text{S}}\left( {{\text{rhombic}}} \right){\text{ + }}\dfrac{3}{2}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{S}}{{\text{O}}_{\text{3}}}\left( {\text{g}} \right)$ , $\Delta {H_1}$
iii).${\text{S}}\left( {{\text{monoclinic}}} \right){\text{ + }}\dfrac{3}{2}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{S}}{{\text{O}}_{\text{3}}}\left( {\text{g}} \right)$ , $\Delta {H_2}$
iii).${\text{S}}\left( {{\text{rhombic}}} \right){\text{ + }}{{\text{O}}_{\text{3}}}\left( {\text{g}} \right) \to {\text{S}}{{\text{O}}_{\text{3}}}\left( {\text{g}} \right)$, $\Delta {H_3}$
iv).${\text{S}}\left( {{\text{monoclinic}}} \right){\text{ + }}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{S}}{{\text{O}}_{\text{3}}}\left( {\text{g}} \right)$ , $\Delta {H_4}$
A.$\Delta {H_1} < \Delta {H_2} < \Delta {H_4}$ (Magnitude)
B.$\Delta {H_1} < \Delta {H_3} < \Delta {H_4}$ (Magnitude)
C.$\Delta {H_1} < \Delta {H_2} = \Delta {H_3} < \Delta {H_4}$ (Magnitude)
D.$\Delta {H_1} + \Delta {H_4} = \Delta {H_2} + \Delta {H_3}$

Answer 1

In periodic table totally $118$ elements. In the periodic table there are a total of seven periods. The period is nothing but the horizontal rows. In the periodic table there are eighteen groups. The group is nothing but the vertical column in the periodic table. This periodic table has some periodic trends in the elements. The atomic radius is one of the factors in this trend. In thermodynamics, enthalpy is an important term. The enthalpy depends on the physical nature of the reactant. The unit and sign of the enthalpy is very important for chemical reaction. If the sign of enthalpy is negative, the reaction is exothermic. The sign of enthalpy is positive, the reaction is endothermic. If the chemical reaction is reversed, we will find out by using signs of enthalpy.

Complete answer:
As we know that the monoclinic sulphur and rhombic sulphur are the types of sulphur. This classification of sulphur is based on the physical appearance of crystals.
The given data is
the reactions:
${\text{S}}\left( {{\text{rhombic}}} \right){\text{ + }}\dfrac{3}{2}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{S}}{{\text{O}}_{\text{3}}}\left( {\text{g}} \right)$ , $\Delta {H_1}$
${\text{S}}\left( {{\text{monoclinic}}} \right){\text{ + }}\dfrac{3}{2}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{S}}{{\text{O}}_{\text{3}}}\left( {\text{g}} \right)$ , $\Delta {H_2}$
${\text{S}}\left( {{\text{rhombic}}} \right){\text{ + }}{{\text{O}}_{\text{3}}}\left( {\text{g}} \right) \to {\text{S}}{{\text{O}}_{\text{3}}}\left( {\text{g}} \right)$, $\Delta {H_3}$
${\text{S}}\left( {{\text{monoclinic}}} \right){\text{ + }}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{S}}{{\text{O}}_{\text{3}}}\left( {\text{g}} \right)$ , $\Delta {H_4}$
Monoclinic sulphur is more stable than rhombic sulphur. Because the structure nature of the monoclinic sulphur is more stable than rhombic sulphur.
And one more thing here, we note the oxygen molecule is more stable than ozone molecule, because bivalency is more stable than trivalency.
According to above discussion, we conclude is the incorrect statement is
$\Delta {H_1} < \Delta {H_2} = \Delta {H_3} < \Delta {H_4}$ (Magnitude).

Hence, option C is the answer.

Note:
We need to know that the sulphur is one of the non-metal in the periodic table. Sulphur is one of the elements in the oxygen family in the periodic table. Sulphur is not soluble in water. But sulphur is soluble in hydrocarbons like benzene and toluene. Sulphur is a yellow colour solid in nature.