Given the following thermo chemical equations (i) S(rhombic) + O2(g) -----> SO2(g), ∆H = - 297.5 kj mol^-1
Given the following thermo chemical equations
(i) S(rhombic) + O2(g) -----> SO2(g), ∆H = - 297.5 kj mol-1
(ii) S(monoclinic) + O2 ------> SO2(g),∆H = - 300.0 kj mol-1
Calculate ∆H for the transformation of one gram atom of rhombic sulphur into monoclinic sulphur.
4 views
1 Answers
Required equation
S(rhombic) ------> S(monoclinic), ∆H = ?
Subtract equation (ii) from (i) it gives
S(rhombic) – S(monoclinic) ----> 0, ∆H =297.5 – (-300.0) = 2.5 KJmol-1
OR
S(rhombic)------> S(monoclinic), ∆H = +2.5 kj mol-1
Thus for the transformation of one gram atom of rhombic sulphur into monoclinic sulphur, 2.5 kj mol-1 of heat is absorbed.
4 views
Answered