Q2 / 90 · JEE Main 2017 (02 Apr) · Chemistry

Given:

Cgraphite+O2g→CO2g ;ΔrHo=−393.5 kJ mol−1

H2g+12O2g→H2Ol;ΔrHo=-285.8 kJ mol-1

CO2g+2H2Ol→CH4g+2O2g;ΔrHo=+890.3 kJ mol-1

Based on the above thermochemical equations, the value of ΔrHo at 298 K for the reaction

Cgraphite+2H2g→CH4g will be:

  • A
    +144.0 kJ mol-1
  • Correct answer
    B
    -74.8 kJ mol-1
  • C
    -144.0 kJ mol-1
  • D
    +74.8 kJ mol-1

Solution

The standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions.

The enthalpy of combustion of a substance is defined as the heat energy given out when one mole of a substance burns completely in oxygen.

The formation reaction of methane is 

Cgraphite+2H2g→CH4g

The heat of combustion data for C(graphite), H2g and CH4g

∆Hfo=∑∆HCombustionreactants-∑∆HCombustionproducts∆HfoCH4=∆HCoCgraphite+2×∆HCoH2g-∆HCoCH4g

∆HfoCH4=-393.5-2×285.8+890.3
ΔfHo of CH4 (g)=−74.8 kJ/mol

Hence, the correct answer is option: 'B'.

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