A) 110.5kJ
B) 676.5 kJ
C) \[-676.5\text{ }kJ\]
D) \[-110.5kJ\]
Correct Answer: D
Solution :
I: \[C(s)+{{O}_{2}}(g)\xrightarrow[{}]{{}}C{{O}_{2}}(g)\]\[\Delta H=-393.5\,kJ\] II: \[CO(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}C{{O}_{2}}(g),\] \[\Delta H=-283.0\,kJ\] \[I-II\] gives III:\[C(s)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}CO(g),\] \[\Delta H=-110.5\,kJ\] This equation III also represents formation of one mol of CO and thus enthalpy change is the heat of formation of CO (g).You need to login to perform this action.
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