Given the following thermochemical equations : (i) S( rhombic ) + O_(2)(g) rarr SO_(2)(g) , Delta H = - 297.5 kJ mol^(-1) (ii) S ( monoclinic) + O_(2) rarrSO_(2)(g) , Delta H = - 300.0 kJ mol^(-1) Calculate Delta Hfor the transformation of one gram atomof rhombicsulphur into monochlinic sulphur.

Given the following thermochemical equations : (i) S( rhombic ) + O_(

1.	Given the following thermochemical equations : (i) S( rhombic ) + O_(2)(g) rarr SO_(2)(g) , Delta H = - 297.5 kJ mol^(-1) (ii) S ( monoclinic) + O_(2) rarrSO_(2)(g) , Delta H = - 300.0 kJ mol^(-1) Calculate Delta Hfor the transformation of one gram atomof rhombicsulphur into monochlinic sulphur.
Answer» Solution :We aim at`:`S ( RHOMBIC ) `rarr ` S ( MONOCLINIC) , `Delta H = ?` EQUATION (i) - Equation(II) gives S ( rhombic ) - S ( monoclinic ) ` rarr 0 , Delta= 297.5 - ( - 300.0) = 2.5 kJ mol^(-1)` or S( rhombic ) `rarr ` S ( monoclinic ) , `Delta H = + 2.5 kJ mol^(-1)` Thus, for the transformation of one gram atom of rhombic sulphur into monoclinicsulphur, `2.5 kJ mol^(-1)` of heat is ABSORBED.