Question

C(s)+O₂(g)→CO₂(g)+400 kJ
C(s)+\(\frac{1}{2}\) O₂(g)→CO(s)+100 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 _______.

• A. 1600 kJ
• B. 3200 kJ
• C. 4400 kJ
• D. 6600 kJ

Answer 1

The Correct Option is D

To calculate the heat generated when burning coal with given conditions, we follow these steps:

1. Identify the reactions:
   • The complete combustion of carbon to form carbon dioxide (CO₂): C(s) + O_2(g) \rightarrow CO_2(g) + 400 \, \text{kJ}
   • The incomplete combustion of carbon to form carbon monoxide (CO): C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g) + 100 \, \text{kJ}
2. Determine the amount of pure carbon in the coal:
   • Given the coal purity is 60%, the mass of pure carbon in 0.6 kg of coal is: 0.6 \, \text{kg} \times 0.60 = 0.36 \, \text{kg}
3. Calculate the distribution of carbon during combustion:
   • 60% of carbon is converted to CO, and 40% is converted to CO₂.
   • Mass of carbon converted to CO: 0.36 \, \text{kg} \times 0.60 = 0.216 \, \text{kg}
   • Mass of carbon converted to CO₂: 0.36 \, \text{kg} \times 0.40 = 0.144 \, \text{kg}
4. Calculate the heat generated:
   • Moles of carbon (C) are calculated by dividing mass by atomic weight of carbon (12 g/mol):
     • Moles of carbon to CO: \frac{0.216 \times 1000}{12} = 18 \, \text{mol}
     • Moles of carbon to CO₂: \frac{0.144 \times 1000}{12} = 12 \, \text{mol}
   • The heat generated from each reaction:
     • Heat from formation of CO: 18 \times 100 = 1800 \, \text{kJ}
     • Heat from formation of CO₂: 12 \times 400 = 4800 \, \text{kJ}
   • Total heat generated: 1800 + 4800 = 6600 \, \text{kJ}
5. Conclusion:
   • The heat generated when 0.6 kg of coal is burnt under the given conditions is 6600 kJ.

Answer 2

Weight of coal = 0.6 kg = 600 gm
∴ 60% of it is carbon
So weight of carbon=600×\(\frac{60}{100}\)=360 g
∴ moles of carbon =\(\frac{360}{12}\)=30 moles
C(12 moles)+O₂⟶CO₂
C(18moles(60% of total carbon)+\(\frac{1}{2}\)O₂⟶CO
∴Heat generated =12×400+18×100
=6600 kJ
So, the correct option is (D): 6600 kJ.