JEE Main · 2022 · Shift-IIhardTHERMO-137

C(s) + O2(g) CO2(g) + 400\,kJ; C(s) + 12O2(g) CO(g) + 100\,kJ When coal of purity 60% is allowed to burn in presence of…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

C(s)+O2(g)CO2(g)+400kJ\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + 400\,\text{kJ}; C(s)+12O2(g)CO(g)+100kJ\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g) + 100\,\text{kJ}

When coal of purity 60% is allowed to burn in presence of insufficient oxygen, 60% of carbon is converted into 'CO' and the remaining is converted into 'CO₂'. The heat generated when 0.6 kg of coal is burnt is:

Options
  1. a

    1600 kJ

  2. b

    3200 kJ

  3. c

    4400 kJ

  4. d

    6600 kJ

Correct Answerd

6600 kJ

Detailed Solution

🧠 Two combustion pathways: scale each by the moles that follow that path Pure carbon content determines total moles. Those moles split between CO and \ceCO2\ce{CO2} formation according to the given percentages, each releasing different heat per mole.

🗺️ Calculation Pure carbon in 0.6 kg of 60% purity coal: 0.6×103×0.60=360g0.6 \times 10^3 \times 0.60 = 360\,\text{g}

n(\ceC)=36012=30moln(\ce{C}) = \frac{360}{12} = 30\,\text{mol}

Forming CO (60%): 18mol×100kJ mol1=1800kJ18\,\text{mol} \times 100\,\text{kJ mol}^{-1} = 1800\,\text{kJ}

Forming \ceCO2\ce{CO2} (40%): 12mol×400kJ mol1=4800kJ12\,\text{mol} \times 400\,\text{kJ mol}^{-1} = 4800\,\text{kJ}

Total = 1800+4800=6600kJ1800 + 4800 = 6600\,\text{kJ}

Answer: (d) 6600 kJ\boxed{\text{Answer: (d) 6600 kJ}}

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C(s) + O2(g) CO2(g) + 400\,kJ; C(s) + 12O2(g) CO(g) + 100\,kJ When coal of purity 60% is allowed to… (JEE Main 2022) | Canvas Classes