Question

Methanol (ch3oh) has been proposed as an alternative fuel. calculate the standard enthalpy of combustion per gram of liquid methanol. standard heats of formation: ch3oh(l) = –239 kj/mol o2(g) = 0 kj/mol co2(g) = –393.5 kj/mol h2o(l) = –286 kj/mol

Answer 1

Answer:The standard enthalpy of combustion per gram of liquid methanol is 45.40kJ/g

Explanation:

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

$\Delta H_{f,CH3OH(l)}=-239 kJ/mol$

$\Delta H_{f,O_2(g)}=0 kJ/mol$

$\Delta H_{f,CO_2(g)}=-393.5 kJ/mol$

$\Delta H_{f,H_2O(l)}=-286 kJ/mol$

Standard enthalpy of combustion of methanol :$\sum{\Delta H_{f,products}}-\sum{\Delta H_{f,reactants}}$

$=(2\times \Delta H_{f,CO_2(l)}+4\times \Delta H_{f,H_2O(g)})-(2\times \Delta H_{f,CH3OH(l)}+3\times \Delta H_{f,O_2(g)})$

$=[2\times (-393.5 kJ/mol)+4\times (-286 kJ/mol)]-[2\times (-239 kJ/mol)+3\times (0 kJ/mol)]=-1453 kj/mol$

The standard enthalpy of combustion per gram of liquid methanol:

$\frac{Delta H_{combustion}}{\text{molar mass of}CH_3OH}=\frac{-1453kJ/mol }{32g/mol}=-45.40kJ/g$

The standard enthalpy of combustion per gram of liquid methanol is 45.40kJ/g

Answer 2

The enthalpy of combustion per gram is -357.9  KJ.

The equation for the combustion of methanol is written as follows;

2CH3OH(l) + 3O2(g)⟶2CO2(g) + 4H2O(l)

From the information in the question, we know that;

ΔHf CH3OH(l) =  –239 KJ/mol

ΔHf O2(g) = 0 KJ/mol

ΔHf CO2(g) =  –393.5 KJ/mol

ΔHf H2O(l) =  –286 KJ/mol

We can calculate the enthalpy of combustion from the heats of formation as follows;

ΔHrxn = ∑ΔHf(products) -  ∑ΔHf(reactants)

ΔHrxn =  [2(–393.5 KJ/mol) + 4( –286 KJ/mol)]  - [2( –239 KJ/mol) + 3(0 KJ/mol)]

ΔHrxn = -11453 KJ/mol

1 mole of methanol contains 32 g

If 32 g of methanol releases 11453 KJ of heat

1 g of methanol releases 11453 KJ  × 1 g /32 g

= -357.9  KJ

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