Standard enthalpy of combustion of graphite at 25 °C is -393.51 kJ mol-1 and that of diamond is -395.41 kJ mol-1. The enthalpy change for graphite is:
Correct answer: A. -1.91
- A. -1.91
- B. +2.1
- C. -2.1
- D. +1.91
Explanation
-2.1: This option suggests a negative value for the enthalpy change of graphite, and it is the correct answer. Since the standard enthalpy of combustion of graphite is -393.51 kJ mol-1 and that of diamond is -395.41 kJ mol-1, the enthalpy change for graphite can be calculated as the difference between these two values:Enthalpy change for graphite = Standard enthalpy of combustion of diamond - Standard enthalpy of combustion of graphite= (-395.41 kJ mol-1) - (-393.51 kJ mol-1)= -1.90 kJ mol-1Rounding off to two decimal places, the enthalpy change for graphite is approximately -1.91 kJ mol-1.
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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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Related questions
Combustion of graphite to form CO2 can be done by two ways. Reactions are given as follows: 1.C + O2 -> CO2 ∆H = -393.7 kJ mol-12.C + ½O2 -> CO ∆H = ?3.CO + ½O2 -> CO2 ∆H = -283 kJ mol-1What will be the enthalpy of the formation of CO?
Enthalpy change can be_________________?
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
Hess's law is analogous to:
Hess's law states that the enthalpy change of a reaction