A) 110.5 kJ
B) 676.5 kJ
C) -676.5 kJ
D) -110.5 kJ
Correct Answer: D
Solution :
I. \[C(s)+{{O}_{2}}(g)\xrightarrow{{}}C{{O}_{2}}(g);\] \[\Delta H=-393.5\,kJ\] II. \[CO(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}C{{O}_{2}}(g);\] \[\Delta H=-283.0\,kJ\] On subtracting Eq. (II) from Eq. (I), we get III. \[C(s)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}CO(g);\] \[\Delta H=-110.5\,kJ\] The equation III also represents formation of 1 mole of CO and thus, enthalpy change, ie, -110.5 kJ, is the heat of formation of \[CO(g)\].
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