`C+O_(2)rarr CO_(2), Delta H = -94 K.cal, CO+(1)/(2)O_(2) rarr CO_(2), Delta H = -67.7 K.cal`. On the basis of the above data, the heat of formation of 'CO' is

1) C_("Graphite") + O_(2(g)) rarr CO_(2(g)) , DeltaH = -94 K.Cals 2) C_("Diamond") + O_(2(g)) rarr CO_(2(g)) , DeltaH =-94.5 K.Cals From the above data the heat of transition of C_("diamond") rarr C_("Graphite")

C_(2)H_(2) + 5/2 O_2(2) rarr 2CO_(2) + H_(2)O , Delta H = -310 kcal C + O_(2) rarr CO_(2) , " "Delta H = -94 kcal H_(2) + 1/2 O_(2) rarr H_(2)O, " " Delta H = -68 kcal On the basis of the above equations, DeltaH_(f) (enthalpy of formation) of C_2H_2 will be :

C_(2)H_(2) + 5/2 O_2(2) rarr 2CO_(2) + H_(2)O , Delta H = -310 kcal C + O_(2) rarr CO_(2) , " "Delta H = -94 kcal H_(2) + 1/2 O_(2) rarr H_(2)O, " " Delta H = -68 kcal On the basis of the above equations, DeltaH_(f) (enthalpy of formation) of C_2H_2 will be :

C + O_(2) rarr CO_(2), CO + (1)/(2) O_(2) rarr CO_(2), Then Delta_(f) H for CO will be

N_(2) + 3H_(2) rarr 2NH_(3), Delta H = - 46K . Cals. From the above reaction, heat of formation of ammonia is

Consider the following reactions. I. C("graphite") + O_(2)(g) rarr CO_(2)(g) , DeltaH^@ = -x_(1)cal II> CO(g) + 1/2 O_(2)(g) rarr CO_(2)(g) , Delta H^(@) = -x_(2) cal Based on the above data, DeltaH_(f)^(@)(CO_(2)) is:

Consider the following reactions. I. C("graphite") + O_(2)(g) rarr CO_(2)(g) , DeltaH^@ = -x_(1)cal II> CO(g) + 1/2 O_(2)(g) rarr CO_(2)(g) , Delta H^(@) = -x_(2) cal Based on the above data, DeltaH_(f)^(@)(CO_(2)) is:

C(s)+O_(2)(g)rarr CO_(2)(g)+94.0 K cal. CO(g)+(1)/(2)O_(2)(g)rarr CO_(2)(g), Delta H=-67.7 K cal. From the above reactions find how much heat (Kcal "mole"^(-1) )would be produced in the following reaction : C(s)+(1)/(2)O_(2)(g)rarr CO(g)

C(s)+O_(2)(g)rarr CO_(2)(g)+94.0 K cal. CO(g)+(1)/(2)O_(2)(g)rarr CO_(2)(g), Delta H=-67.7 K cal. From the above reactions find how much heat (Kcal "mole"^(-1) )would be produced in the following reaction : C(s)+(1)/(2)O_(2)(g)rarr CO(g)