Question

Hydrogen and Methanol have both been proposed as alternatives to hydrocarbon fuels. Write balanced reactions for the complete combustion of hydrogen. Use standard enthalpies of formation to calculate the amount of heat released per kilogram of the fuel.

Answer 1

Answer: The amount of heat released per kilogram of hydrogen is -142,750 kJ and per kilogram of methanol is -22690.3 kJ

Step-by-step explanation:

• When fuel is hydrogen:

The chemical equation for the combustion of hydrogen follows:

`H_2(g)+(1)/(2)O_2(g)\rightarrow H_2O(l)`

The equation used to calculate enthalpy change is of a reaction is:

`\Delta H^o_(rxn)=\sum [n* \Delta H^o_f_((product))]-\sum [n* \Delta H^o_f_((reactant))]`

The equation for the enthalpy change of the above reaction is:

`\Delta H^o_(rxn)=[(1* \Delta H^o_f_((H_2O(l))))]-[(1* \Delta H^o_f_((H_2(g))))+((1)/(2)* \Delta H^o_f_((O_2(g))))]`

We are given:

`\Delta H^o_f_((H_2O(l)))=-285.8kJ/mol\\\Delta H^o_f_((H_2(g)))=0kJ/mol\\\Delta H^o_f_((O_2(g)))=0kJ/mol`

Putting values in above equation, we get:

`\Delta H^o_(rxn)=[(1* (-285.8))]-[(1* 0)+((1)/(2)* 0)]=-285.8kJ/mol`

Mass of hydrogen reacted = 1 mole = 2 grams

Conversion factor used: 1 kg = 1000 g

Applying unitary method:

When 2 grams of hydrogen is combusted, the amount of heat released is 285.8 kJ

So, when 1000 grams of hydrogen is combusted, the amount of heat released will be
`(285.5)/(2)* 1000=142,750kJ`

Amount of heat released per kilogram of hydrogen combusted = -142,750 kJ

• When fuel is methanol:

The chemical equation for the combustion of hydrogen follows:

`2CH_3OH(l)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)`

The equation for the enthalpy change of the above reaction is:

`\Delta H^o_(rxn)=[(2* \Delta H^o_f_((CO_2(g))))+(4* \Delta H^o_f_((H_2O(l))))]-[(2* \Delta H^o_f_((CH_3OH(l))))+(3* \Delta H^o_f_((O_2(g))))]`

We are given:

`\Delta H^o_f_((CH_3OH(l)))=-238.42kJ/mol\\\Delta H^o_f_((CO_2(g)))=-393.51kJ/mol\\\Delta H^o_f_((H_2O(l)))=-285.5kJ/mol\\\Delta H^o_f_((O_2(g)))=0kJ/mol`

Putting values in above equation, we get:

`\Delta H^o_(rxn)=[(2* (-393.51))+(4* (-285.5))]-[(2* (-238.42))+(3* 0)]=-1452.18kJ/mol`

Mass of methanol reacted = 2 moles = 64 grams

Applying unitary method:

When 64 grams of methanol is combusted, the amount of heat released is 1452.18 kJ

So, when 1000 grams of methanol will be combusted, the amount of heat released will be
`(1452.18)/(64)* 1000=22690.3kJ`

Amount of heat released per kilogram of methanol combusted = -22690.3 kJ