If the sublimation energy and enthalpy of fusion of I 2 are 57.3 kJ mol -1 and 15.5 kJ mol -1, respectively then, the enthalpy of vaporisation of I 2 will be

Question

If the sublimation energy and enthalpy of fusion of I 2 are 57.3 kJ mol -1 and 15.5 kJ mol -1, respectively then, the enthalpy of vaporisation of I 2 will be (A) -72.8 kJ mol -1 (B) 72.8 kJ mol -1 (C) -41.8 kJ mol -1 (D) +41.8 kJ mol -1

Tardigrade Admin · Accepted Answer

(i) I 2(s) longrightarrow I 2(g) ; Δ H1=+57.3 kJ mol -1 (ii) I 2(s) longrightarrow I 2(l) ; Δ H2=+15.5 kJ mol -1 (iii) I 2(l) longrightarrow I 2(g) ; Δ H= ? Eq (iii) will be obtained, if we substract Eq (ii) from (i), i.e (iii) = (i) - (ii) Now, Δ H=57.3-15.5=+41.8 kJ mol -1

Q. If the sublimation energy and enthalpy of fusion of $I_{2}$ are $57.3 k J m o l^{- 1}$ and $15.5 k J m o l^{- 1}$ , respectively then, the enthalpy of vaporisation of $I_{2}$ will be

$- 72.8 k J m o l^{- 1}$

$72.8 k J m o l^{- 1}$

$- 41.8 k J m o l^{- 1}$

$+ 41.8 k J m o l^{- 1}$

Q. If the sublimation energy and enthalpy of fusion of I2​ are 57.3kJmol−1 and 15.5kJmol−1, respectively then, the enthalpy of vaporisation of I2​ will be

−72.8kJmol−1

72.8kJmol−1

−41.8kJmol−1

+41.8kJmol−1

(i) I2​(s)⟶I2​(g);ΔH1​=+57.3kJmol−1 (ii) I2​(s)⟶I2​(l);ΔH2​=+15.5kJmol−1 (iii) I2​(l)⟶I2​(g);ΔH= ? Eq (iii) will be obtained, if we substract Eq (ii) from (i), i.e (iii) = (i) - (ii) Now, ΔH=57.3−15.5=+41.8kJmol−1

Q. If the sublimation energy and enthalpy of fusion of $I_{2}$ are $57.3 k J m o l^{- 1}$ and $15.5 k J m o l^{- 1}$ , respectively then, the enthalpy of vaporisation of $I_{2}$ will be

$- 72.8 k J m o l^{- 1}$

$72.8 k J m o l^{- 1}$

$- 41.8 k J m o l^{- 1}$

$+ 41.8 k J m o l^{- 1}$