propyne, propylene, and propane

logistic_guy

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The standard enthalpies of formation of gaseous propyne (C3H4)\displaystyle (C_3H_4), propylene (C3H6)\displaystyle (C_3H_6), and propane (C3H8)\displaystyle (C_3H_8) are +185.4,+20.4,\displaystyle +185.4, +20.4, and 103.8\displaystyle -103.8 kJ/mol, respectively. (a)\displaystyle \left(\bold{a}\right) Calculate the heat evolved per mole on combustion of each substance to yield CO2(g)\displaystyle CO_2(g) and H2O(g)\displaystyle H_2O(g). (b)\displaystyle \left(\bold{b}\right) Calculate the heat evolved on combustion of 1\displaystyle 1 kg of each substance. (c)\displaystyle \left(\bold{c}\right) Which is the most efficient fuel in terms of heat evolved per unit mass?
 
The standard enthalpies of formation of gaseous propyne (C3H4)\displaystyle (C_3H_4), propylene (C3H6)\displaystyle (C_3H_6), and propane (C3H8)\displaystyle (C_3H_8) are +185.4,+20.4,\displaystyle +185.4, +20.4, and 103.8\displaystyle -103.8 kJ/mol, respectively. (a)\displaystyle \left(\bold{a}\right) Calculate the heat evolved per mole on combustion of each substance to yield CO2(g)\displaystyle CO_2(g) and H2O(g)\displaystyle H_2O(g). (b)\displaystyle \left(\bold{b}\right) Calculate the heat evolved on combustion of 1\displaystyle 1 kg of each substance. (c)\displaystyle \left(\bold{c}\right) Which is the most efficient fuel in terms of heat evolved per unit mass?
show us your effort/s to solve this problem.
 
show us your effort/s to solve this problem.
👍

Let me try to solve (a) for propyne. The combustion reaction is:

C3H4(g)+4O2(g)3CO2(g)+2H2O(g)\displaystyle \text{C}_3\text{H}_4(g) + 4\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 2\text{H}_2\text{O}(g)

The heat per mole is:

ΔH=nΔHf(products)mΔHf(reactants)\displaystyle \Delta H = \sum n\Delta H^{\circ}_f(\text{products}) - \sum m\Delta H^{\circ}_f(\text{reactants})

where ΔHf\displaystyle \Delta H^{\circ}_f is the standard enthalpy of formation.

ΔH=[3ΔHf(CO2)+2ΔHf(H2O)][ΔHf(C3H4)+4ΔHf(O2)]\displaystyle \Delta H = [3\Delta H^{\circ}_f(\text{CO}_2) + 2\Delta H^{\circ}_f(\text{H}_2\text{O})] - [\Delta H^{\circ}_f(\text{C}_3\text{H}_4) + 4\Delta H^{\circ}_f(\text{O}_2)]

The standard enthalpy of formation for propyne was given, but for carbon dioxide and water, I have to look at the table.

enthalpy.gif
The standrard enthalpy of formation of any element in its most stable form such as O2\displaystyle \text{O}_2 is defined to be zero.

ΔH=[3(393.5)+2(241.8)][(185.4)+4(0)]=1849.5 kJ/mol\displaystyle \Delta H = [3(-393.5) + 2(-241.8)] - [(185.4) + 4(0)] = -1849.5 \ \text{kJ/mol}

I will continue in the next post.
 
Let us start with the combustion reaction for propylene.

C3H6(g)+4.5O2(g)3CO2(g)+3H2O(g)\displaystyle \text{C}_{3}\text{H}_{6}(g) + 4.5\text{O}_{2}(g) \rightarrow 3\text{C}\text{O}_{2}(g) + 3\text{H}_{2}\text{O}(g)

💪😭
 
(b) propylene\displaystyle \bold{(b)} \rightarrow \ \text{propylene}

ΔH=[3ΔHf(CO2)+3ΔHf(H2O)][ΔHf(C3H6)+4.5ΔHf(O2)]\displaystyle \Delta H = [3\Delta H^{\circ}_f(\text{CO}_2) + 3\Delta H^{\circ}_f(\text{H}_2\text{O})] - [\Delta H^{\circ}_f(\text{C}_3\text{H}_6) + 4.5\Delta H^{\circ}_f(\text{O}_2)]

=[3(393.5)+3(241.8)][(20.4)+4.5(0)]=1926.3 kJ/mol\displaystyle = [3(-393.5) + 3(-241.8)] - [(20.4) + 4.5(0)] = -1926.3 \ \text{kJ/mol}
 
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