chemistry part 1 class 11 pg no.183 q no 6.14 please can u solve this i am not getting how to do it Share with your friends Share 0 Vartika Jain answered this Dear Student, The reactions that takes place during the formation of CH3OH(l) can be written asC(s) + 2H2O(g) +12O2(g) → CH3OH (l) ..(i)Also, CH3OH(l) +32O2(g) → CO2 + 2H2O(l) ∆H°r ...(ii)C(g) +O2(g) → CO2 ∆H°c ...(iii)H2(g) +12O2(g) → 2H2O(l) ∆H°f ...(iv)(i) can be obtained by =(ii) +2×(iii)-(i)∆H°f(CH3OH)= ∆H°c+2∆H°f(H2O)-∆H°r=(-393)+2(-286)-(-726)=(-393-572+726) kJmol-1or, ∆H°f(CH3OH)=-239 kJmol-1 0 View Full Answer