InterviewSolution
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InterviewSolution
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Compute the heat of formation of liquid methyl alcohol in `kJ mol^(-1)` , using the following `:` Heat of vaporisation of liquid methyl alcohol `=38 kJ mol^(-1)`. Heat of formation of gaseous atoms from the elements in their standard states `: H, 218 kJ//mol , C,715 kJ // mol, O, 249 kJ //mol`. Averagebond energies `: C-H, 415 kJ //mol ,C-O kJ //mol, O-H, 463 kJ //mol` |
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Answer» We aim at `: C(s) +2H_(2)(g) + (1)/(2)O_(2) rarr CH_(3)OH(l), DeltaH =?` We are given `: (i) CH_(3)OH(l) rarr CH_(3) OH (g), DeltaH = 38 kJ mol^(-1)` (iii) `C(s)rarr C(g), DeltaH =715 kJ mol^(-1)` (iv) `(1)/(2)O_(2)(g) rarrO(g) , DeltaaH -249 kJ mol^(-1)` Also from the given bond energies, we have (v) `CH_(3)OH(g)(H- underset(H) underset(|) overset(H)overset(|) (C) -O-H) rarr C(g) + 4H(g)+O(g), DeltaH = 3 xx 415 +356 +463 = 2064kJ mol^(-1)` Eqn. (iii) `+4 ` Eqn. (ii) `+` Eqn. (iv)- Eqn.(i) - Eqn. (v) gives the requried result i.e., `DeltaH = 715 + 4( 218) + 249-38 -2064 = -266 kJ mol^(-1)` |
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