Question 14.2: Chemical reaction, C(s) + O2 (g) → CO2 (g) at 25°C and 1 atm......
Chemical reaction, C(s) + O_2 (g) → CO_2 (g) at 25°C and 1 atm releases –17.9 kcal of heat, if temperature of outlet, i.e. CO_2 is kept constant at 25°C and 1 atm respectively.
Calculate the enthalpy of formation for CO_2.
Using the value of enthalpy of formation for CO_2, find out the enthalpy of combustion of Methane, which is burnt in presence of pure oxygen. Given here:
ΔH^0_{CH_4} = -94.0 kCal ; ΔH^0_{O_2} = -26.3 kcal ; ΔH^0_{CO_2} = ? ; ΔH^0_{H_2O} = 0.0 kcal
Given,
1. For CO_2 formation, heat released, Q = –17.9 kcal
2. Enthalpy change data
Find out,
1. Enthalpy of formation for CO_2
2. Enthalpy of combustion for methane.
Analysis,
(a) From the definition of enthalpy of formation, ΔH^0_{CO_2} for CO_2 = Q = –17.9 kcal.
(b) Chemical reaction for the burning of Methane is:
CH_4 (g) + 2O_2 (g)\longrightarrow CO_2 (g) + 2H_2O (l)
Enthalpy of combustion = Sum of enthalpies of all reactants – Sum of enthalpies of all products
H_{RP} = \sum{n_P}(H^0_f + \Delta H) – \sum{n_R}(H^0_f + \Delta H)
or
H_{RP} = \left[n_{CO2} \times (H^0_f + \Delta H)_{CO_2}) + n_{H_2O} \times (H^0_f + \Delta H)_{H_2o}\right] \\- \left[n_{CH4} \times ((H^0_f + \Delta H)_{CH_4} + n_{o_2} \times (H^0_f + \Delta H)_{o_2})\right]
Since reaction takes place at reference temperature (for both reactants and products), terms ΔH = 0.
H_{RP} = \left[(1x – 94.0) + (2x – 26.3)\right] – \left[(1x – 17.19) + (2 \times 0)\right] = – 212.70 kCal