Combustion of graphite to form CO2 can be done in two ways. Reactions are given as follows:
Correct answer: B. -110 kJ mol-1
- A. -676 kJ mol-1
- B. -110 kJ mol-1
- C. 110 kJ mol-1
- D. 676 kJ mol-1
Explanation
To find the enthalpy change for the combustion of graphite to form CO, apply Hess's Law by using the given reactions. According to Hess's Law, the total enthalpy change is independent of the path taken. Thus, set up the equation: ΔH CO (eqn 2) + ΔH CO2 (eqn 3) = ΔH CO2 (eqn 1). Substitute the known values: ΔH CO + (-283) = -393.7. Solving for ΔH CO gives ΔH CO = -393.7 + 283, resulting in ΔH CO = -110 kJ mol-1. Option B is correct. Options A and D provide values that are not derived from the given reaction data, while Option C ignores the necessary negative sign for exothermic reactions.
Last updated
About Thermochemistry and Energetics of Chemical Reactions
Thermochemistry measures energy changes in chemical reactions and distinguishes exothermic from endothermic processes. Work covers systems, surroundings and state functions, internal energy, the first law of thermodynamics, enthalpy and Hess's law, including the sign conventions used when heat enters or leaves a system.
Practise Thermochemistry and Energetics of Chemical Reactions
728 free Thermochemistry and Energetics of Chemical Reactions MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Chemistry questions like this
Chemistry is on 12 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
1/2 H2(g) ➞ H(g) ∆H = 218 kJ mol-1.In this reaction, ∆H will be called:
1 Kcal is equal to:
2P(s) 3CI2(g) 2PCI3 ∆H= 151.8 kJ PCl3+ CI2(g) , PCI5(g); ∆H = -32.8 kJFrom the following data, the heat of formation of PCI5 comes out to be :
5 calories are equivalent to _ Joule.
5 mole of an ideal gas expand reversibly from a volume of 8 dm3 to 80 dm3 at a temperature of 27°C. Calculate the change in entropy.