Question

Many portable gas heaters and grills use propane, C3H8(g). Using enthalpies of formation, calculate the quantity of heat produced when 13.0 g of propane is completely combusted in air under standard conditions. Assume that liquid water is forming. Express the heat in kilojoules to three significant digits.

Answer 1

Answer: The amount of heat released is -603. kJ

Step-by-step explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

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

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

The chemical equation for the combustion of propane follows:

`C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)`

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

`\Delta H_(rxn)=[(3* \Delta H_f_((CO_2(g))))+(4* \Delta H_f_((H_2O(g))))]-[(1* \Delta H_f_((C_3H_8(g))))+(5* \Delta H_f_((O_2(g))))]`

We are given:

`\Delta H_f_((H_2O(g)))=-241.82kJ/mol\\\Delta H_f_((CO_2(g)))=-393.51kJ/mol\\\Delta H_f_((C_3H_8(g)))=-103.8kJ/mol\\\Delta H_f_((O_2(g)))=0kJ/mol`

Putting values in above equation, we get:

`\Delta H_(rxn)=[(3* (-393.51))+(4* (-241.82))]-[(1* (-103.8))+(3* (0))]\\\\\Delta H_(rxn)=-2044.01kJ/mol`

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Given mass of propane = 13.0 g

Molar mass of propane = 44 g/mol

Putting values in above equation, we get:

`\text{Moles of propane}=(13g)/(44g/mol)=0.295mol`

To calculate the enthalpy change of the reaction, we use the equation:

`\Delta H_(rxn)=(q)/(n)`

where,

q = amount of heat released = ?

n = number of moles of propane = 0.295 moles

`\Delta H_(rxn)` = enthalpy change of the reaction = -2044.01 kJ/mol

Putting values in above equation, we get:

`-2044.01kJ/mol=(q)/(0.295mol)\\\\q=(-2044.01kJ/mol* 0.295mol)=-602.98kJ`

Hence, the amount of heat released is -603. kJ