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NEET CHEMISTRYThermodynamicsMedium

Question

The heat of combustion of carbon to CO2CO_2 is -393.5 kJ/mol. The heat released upon the formation of 35.2 g of CO2CO_2 from carbon and oxygen gas is:

A

-315 kJ

B

+315 kJ

C

-630 kJ

D

-3.15 kJ

Step-by-Step Solution

The combustion of carbon to form CO2CO_2 is given by the equation: C(s)+O2(g)CO2(g)C(s) + O_2(g) \rightarrow CO_2(g) For 1 mole of CO2CO_2 (molar mass = 44 g/mol), the heat of combustion (ΔH\Delta H) is 393.5 kJ/mol-393.5 \text{ kJ/mol}. The number of moles in 35.2 g of CO2CO_2 is: n=Given massMolar mass=35.2 g44 g/mol=0.8 molesn = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{35.2 \text{ g}}{44 \text{ g/mol}} = 0.8 \text{ moles} The heat released for the formation of 0.8 moles of CO2CO_2 is: ΔH=0.8 moles×(393.5 kJ/mol)=314.8 kJ315 kJ\Delta H = 0.8 \text{ moles} \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.

CHEMISTRYThermodynamicscombustioncarbonreleasedformationcarbon

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