Answer
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Hint:This question is based on the thermodynamic concept. The heat conversion can be calculated by writing the chemical reactions of the conversion of rhombic sulfur to monoclinic sulfur. The sulfur dioxide will be the product formed in both the reactions.
Complete answer:
->When we heat sulfur in presence of oxygen, sulfur-dioxide is formed. There are two allotropes of sulfur. One is monoclinic and the other one is rhombic. Both of them have different heat of combustion.
\[\begin{matrix}
\text{S }+\text{ }{{\text{O}}_{2}}\to & \text{S}{{\text{O}}_{2}} \\
\text{Sulfur Oxygen } & Sulfur-dioxide \\
\end{matrix}\]
->It is an application of Hess law whose standard enthalpy of conversion can be calculated using the equation given below
\[\Delta {{H}_{conversion}}=\Delta {{H}_{rhombic}}-\Delta {{H}_{monoclinic}}\]
->First, let us write the chemical reaction and as per the question the heat of combustion is also mentioned i.e.
For rhombic sulfur
S$_{R}$ + O$_2$ $\rightleftharpoons$ SO$_2$, $\delta$H$_{R}$ = -70960 cal
For monoclinic sulphur
S$_{M}$ + O$_2$ $\rightleftharpoons$ SO$_2$, $\delta$H$_{M}$= -71030 cal
->In these chemical reactions, $\delta$H$_{R}$, and $\delta$H$_{M}$ represents the heat of combustion for both the reactions respectively.
-Now, we have to find the heat of conversion for rhombic sulfur to monoclinic sulfur, consider it to be $\delta$H$_{req}$
->We have, S$_{R}$ $\rightarrow$ S$_{M}$, we attained it by subtracting the equation of monoclinic sulfur from that of rhombic sulfur.
-Now, $\delta$H$_{req}$ = $\delta$H$_{R}$- $\delta$H$_{M}$
$\delta$H$_{req}$ = -70960- (-71030) cal = +70 cal
->Therefore, we can conclude that the heat of conversion is equal to +70cal.
Hence the correct answer is option ‘c’.
Additional information:It is known to us that after oxygen, sulfur is the second element to have allotropes. Monoclinic sulfur is a crystalline allotrope of sulfur obtained by heating the rhombic sulfur.
When it is left at room temperature, it again reverts back to rhombic sulfur, having presence of ring molecules in the crystalline structure.
Note::There is always confusion between why a negative sign is placed before the value of heat of combustion, or any kind of heat. The negative sign indicates that heat is evolved in this reaction. This is a simple thermodynamic question. Just write the chemical equations, and subtract it.
Complete answer:
->When we heat sulfur in presence of oxygen, sulfur-dioxide is formed. There are two allotropes of sulfur. One is monoclinic and the other one is rhombic. Both of them have different heat of combustion.
\[\begin{matrix}
\text{S }+\text{ }{{\text{O}}_{2}}\to & \text{S}{{\text{O}}_{2}} \\
\text{Sulfur Oxygen } & Sulfur-dioxide \\
\end{matrix}\]
->It is an application of Hess law whose standard enthalpy of conversion can be calculated using the equation given below
\[\Delta {{H}_{conversion}}=\Delta {{H}_{rhombic}}-\Delta {{H}_{monoclinic}}\]
->First, let us write the chemical reaction and as per the question the heat of combustion is also mentioned i.e.
For rhombic sulfur
S$_{R}$ + O$_2$ $\rightleftharpoons$ SO$_2$, $\delta$H$_{R}$ = -70960 cal
For monoclinic sulphur
S$_{M}$ + O$_2$ $\rightleftharpoons$ SO$_2$, $\delta$H$_{M}$= -71030 cal
->In these chemical reactions, $\delta$H$_{R}$, and $\delta$H$_{M}$ represents the heat of combustion for both the reactions respectively.
-Now, we have to find the heat of conversion for rhombic sulfur to monoclinic sulfur, consider it to be $\delta$H$_{req}$
->We have, S$_{R}$ $\rightarrow$ S$_{M}$, we attained it by subtracting the equation of monoclinic sulfur from that of rhombic sulfur.
-Now, $\delta$H$_{req}$ = $\delta$H$_{R}$- $\delta$H$_{M}$
$\delta$H$_{req}$ = -70960- (-71030) cal = +70 cal
->Therefore, we can conclude that the heat of conversion is equal to +70cal.
Hence the correct answer is option ‘c’.
Additional information:It is known to us that after oxygen, sulfur is the second element to have allotropes. Monoclinic sulfur is a crystalline allotrope of sulfur obtained by heating the rhombic sulfur.
When it is left at room temperature, it again reverts back to rhombic sulfur, having presence of ring molecules in the crystalline structure.
Note::There is always confusion between why a negative sign is placed before the value of heat of combustion, or any kind of heat. The negative sign indicates that heat is evolved in this reaction. This is a simple thermodynamic question. Just write the chemical equations, and subtract it.
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