Calculate enthalpy of formation of methane( CH_(4)) from the following data : (I) C (s) + O_(2)(g) rarr CO_(2)(g), Delta _(r)H^(@)= -393.5 kJ mol^(_1) (ii)H_(2)(g) +(1)/(2) O_(2)(g) rarr H_(2)O(l), Delta _(r) H^(@) = - 285.8 kJ mol^(_1) (iii)CH_(4)(g) +2O_(2)(g) rarr CO_(2)(g) +2H_(2)O(l) , Delta _(r) H^(@)= - 890.3 kJ mol^(-1)

Calculate enthalpy of formation of methane( CH_(4)) from the following

1.	Calculate enthalpy of formation of methane( CH_(4)) from the following data : (I) C (s) + O_(2)(g) rarr CO_(2)(g), Delta _(r)H^(@)= -393.5 kJ mol^(_1) (ii)H_(2)(g) +(1)/(2) O_(2)(g) rarr H_(2)O(l), Delta _(r) H^(@) = - 285.8 kJ mol^(_1) (iii)CH_(4)(g) +2O_(2)(g) rarr CO_(2)(g) +2H_(2)O(l) , Delta _(r) H^(@)= - 890.3 kJ mol^(-1)
Answer» Solution :We aimat at `: C(s) + 2H_(2)(g) RARR CH_(4)(g) , Delta_(f)H^(@) = ?` Multiplyingeqn. (ii) with 2, adding to eqn. (i) and then subtracting eqn. (iii) from the sum, i.e., operating eqn. (i) `+2 xx ` eqn. (ii) - eqn. (iii) , we get `C(s) + 2H_(2)(g) - CH_(4)(g) rarr 0,DELTA _(R) H^(@) - 393.5 + 2( -285.8) - ( - 890.3) = - 74.8 kJ mol^(_1)` or `C_(s) + 2H_(2)(g) rarr CH_(4)(g), Delta _(f) H^(@) = -74.8 kJ mol^(-1)` Hence, enthalpy of FORMATION of methane is `: Delta_(f) H^(@) = - 74.8 kJ mol^(-1)`