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MHT CET 202621 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
21 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

MHT CET 2026 · 21 April · Evening Shift · Chemistry · Thermodynamics (C)

Enthalpy of combustion of carbon to carbon dioxide is -390\ kJ/mol...

Enthalpy of combustion of carbon to carbon dioxide is −390 kJ/mol-390\ \text{kJ/mol}. What is the amount of heat released when 35 g35\ \text{g} of carbon dioxide is formed from the reaction of carbon and dioxygen gas?

Options

  1. A245 kJ245\ \text{kJ}
  2. B310 kJ310\ \text{kJ}
  3. C490 kJ490\ \text{kJ}
  4. D700 kJ700\ \text{kJ}

Correct answer

B. 310 kJ310\ \text{kJ}

Step-by-step solution

The combustion reaction of carbon is given by:

C(s)+O2(g)→CO2(g)C(s) + O_2(g) \rightarrow CO_2(g)

The enthalpy of combustion is ΔH=−390 kJ/mol\Delta H = -390\ \text{kJ/mol}. This means 390 kJ390\ \text{kJ} of heat is released when 1 mol1\ \text{mol} of CO2CO_2 is formed.

Molar mass of CO2=12+2×16=44 g/molCO_2 = 12 + 2 \times 16 = 44\ \text{g/mol}

Number of moles in 35 g35\ \text{g} of CO2=3544 molCO_2 = \dfrac{35}{44}\ \text{mol}

Amount of heat released =3544×390 kJ=310.22 kJ≈310 kJ= \dfrac{35}{44} \times 390\ \text{kJ} = 310.22\ \text{kJ} \approx 310\ \text{kJ}

Answer: 310 kJ310\ \text{kJ}
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