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NEET CHEMISTRYThermodynamicsMedium

Question

The heat of combustion of carbon to CO2\text{CO}_2 is −393.5 kJ/mol-393.5 \text{ kJ/mol}. The heat released upon the formation of 35.2 g35.2 \text{ g} of CO2\text{CO}_2 from carbon and oxygen gas is:

A

−315 kJ-315 \text{ kJ}

B

+315 kJ+315 \text{ kJ}

C

−630 kJ-630 \text{ kJ}

D

+630 kJ+630 \text{ kJ}

Step-by-Step Solution

The combustion of carbon to form carbon dioxide is given by the reaction: C(s)+O2(g)→CO2(g);ΔH=−393.5 kJ/mol\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}; \Delta H = -393.5 \text{ kJ/mol} This means that the formation of 1 mole1 \text{ mole} of CO2\text{CO}_2 (44 g44 \text{ g}) releases 393.5 kJ393.5 \text{ kJ} of heat (or the enthalpy change is −393.5 kJ-393.5 \text{ kJ}). Number of moles in 35.2 g35.2 \text{ g} of CO2\text{CO}_2 = Given massMolar mass=35.2 g44 g/mol=0.8 mol\frac{\text{Given mass}}{\text{Molar mass}} = \frac{35.2 \text{ g}}{44 \text{ g/mol}} = 0.8 \text{ mol}. The heat released (enthalpy change) for the formation of 0.8 mol0.8 \text{ mol} of CO2\text{CO}_2 is: ΔH=0.8 mol×(−393.5 kJ/mol)=−314.8 kJ≈−315 kJ\Delta H = 0.8 \text{ mol} \times (-393.5 \text{ kJ/mol}) = -314.8 \text{ kJ} \approx -315 \text{ kJ}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicscombustioncarbontextcoreleasedformation

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