Given that C + O2 gives CO2 has an enthalpy change of minus 393 kJ per mole, and CO + half O2 gives CO2 has an enthalpy change of minus 283 kJ per mole, the enthalpy of formation of carbon monoxide is
Correct answer: C. minus 110 kJ per mole
- A. minus 676 kJ per mole
- B. plus 110 kJ per mole
- C. minus 110 kJ per mole
- D. minus 283 kJ per mole
Explanation
Forming carbon monoxide and then burning it must total the same as burning carbon directly, so the unknown equals minus 393 minus the value of minus 283, giving minus 110 kJ per mole. Adding the two values instead of subtracting gives minus 676, which is the trap. This calculation is the standard textbook illustration of Hess's law.
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About Hess's Law
Hess's law states that the enthalpy change of a reaction is independent of the route because enthalpy is a state function. Calculations involve reversing and adding thermochemical equations, using standard enthalpies of formation or combustion, and finding an unknown reaction enthalpy without confusing enthalpy change with reaction rate.
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Hess's law states that the enthalpy change of a reaction
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