Question

1) The heat of combustion for the gases hydrogen, methane and ethane are −285.8, −890.4 and −1559.9 kJ/mol respectively at 298K.

equation 1 H2 + 1⁄2O2 ---> H2O ΔH = −285.8 kJ
equation 2 CH4 + 2O2 ---> CO2 + 2H2O ΔH = −890.4 kJ
equation 3 C2H6 + 7⁄2O2 ---> 2CO2 + 3H2O ΔH = −1559.9 kJ

Use the above equations to calculate (at the same temperature) the heat of reaction for the following reaction:
2CH4(g) ---> C2H6(g) + H2(g)
Solution:

Answer 1

The enthalpy of the reaction is 64.9 kJ/mol.

Step-by-step explanation:

`H_2 + (1)/(2)O_2\rightarrow H_2O,\Delta H_1 =-285.8 kJ`..[1]

`CH_4 + 2O_2\rightarrow CO_2 + 2H_2O,\Delta H_2 =-890.4 kJ`..[2]

`C_2H_6 + (7)/(2)O_2\rightarrow 2CO_2 + 3H_2O,\Delta H_3= -1559.9 kJ`..[3]

`2CH_4(g)\rightarrow C_2H_6(g) + H_2(g),\Delta H_4=?`..[4]

2 × [2] - [1]- [3] = [4] (Using Hess's law)

`\Delta H_4=2* \Delta H_2 -\Delta H_1 -\Delta H_3`

`\Delta H_4=2* (-890.4 kJ)-(-285.8 kJ) -(-1559.9 kJ)`

`\Delta H_4=64.9 kJ/mol`

The enthalpy of the reaction is 64.9 kJ/mol.