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

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+2H°f(H2O)-H°r=(-393)+2(-286)-(-726)=(-393-572+726) kJmol-1or, H°f(CH3OH)=-239 kJmol-1

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