Question #59876

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Expert's answer

2016-05-13T06:37:09-0400

Answer on Question#59876 - Chemistry | General Chemistry

A) calculate the missing ΔH\Delta H values (using Hess's Law) in the table below: use ΔHfCO2(g)=−394 kj mol−1\Delta H_{\mathrm{f}} \mathrm{CO}_{2}(\mathrm{g}) = -394 \mathrm{~kj} \mathrm{~mol}^{-1} AND ΔHfH2O(I)=−286 kj mol−1\Delta H_{\mathrm{f}} H_{2} O(I) = -286 \mathrm{~kj} \mathrm{~mol}^{-1}


B) use the data to calculate ΔH\Delta H for the reaction


CH3COOH(I)+C2H5OH(I)⟶CH3COOC2H5(I)+H2O(I)\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{I}) + \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{I}) \longrightarrow \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(\mathrm{I}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{I})

Answers:

A) The equation we want for the formation of methane from its elements would be the following:


C(s)+2H2(g)→CH4(g)→CH3COOH=?\mathrm{C}(\mathrm{s}) + 2 \mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{CH}_{4}(\mathrm{g}) \rightarrow \mathrm{CH}_{3} \mathrm{COOH} = ?


You are given the three equations:

1. C(s)+O2(g)→CO2(g)\mathrm{C}(\mathrm{s}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}). ΔH1=−394 kJmol−1\Delta H_{1} = -394 \mathrm{~kJmol}^{-1}

2. H2(g)+1/2O2(g)→H2O(g)\mathrm{H}_{2}(\mathrm{g}) + 1/2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2}\mathrm{O}(\mathrm{g}). ΔH2=−286 kJmol−1\Delta H_{2} = -286 \mathrm{~kJmol}^{-1}

3. CH4(g)+2O2(g)→CO2(g)+2H2O(g)\mathrm{CH}_{4}(\mathrm{g}) + 2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) + 2 \mathrm{H}_{2}\mathrm{O}(\mathrm{g}). ΔH3=−890 kJmol−1\Delta H_{3} = -890 \mathrm{~kJmol}^{-1}

Equation 1 already has carbon on the left side. Equation 2 already has hydrogen on the left side, but we need two hydrogen molecules, so we can double the second equation. Equation 3 has methane on the left side instead of on the right side, so we need to reverse it.

1. C(s)+O2(g)→CO2(g)\mathrm{C}(\mathrm{s}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}). ΔH1=−394 kJmol−1\Delta H_{1} = -394 \mathrm{~kJmol}^{-1}

2. 2H2(g)+O2(g)→2H2O(g)2 \mathrm{H}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2}\mathrm{O}(\mathrm{g}). ΔH2=−572 kJmol−1\Delta H_{2} = -572 \mathrm{~kJmol}^{-1}

3. CO2(g)+2H2O(g)→CH4(g)+2O2(g)\mathrm{CO}_{2}(\mathrm{g}) + 2 \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CH}_{4}(\mathrm{g}) + 2 \mathrm{O}_{2}(\mathrm{g}). ΔH3=+890 kJmol−1\Delta H_{3} = +890 \mathrm{~kJmol}^{-1}

Eliminating oxygen, carbon dioxide, and water on both sides in all the three equations, we end up with:

Total: C(s)+2H2(g)→CH4(g)\mathrm{C}(\mathrm{s}) + 2 \mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{CH}_{4}(\mathrm{g}). ΔH=−76 kJmol−1\Delta H = -76 \mathrm{~kJmol}^{-1}

The standard heat of formation for methane is −76 kJmol−1-76 \mathrm{~kJmol}^{-1}.

Answer: ΔHf(CH4)=−76 kJmol−1\Delta H_{\mathrm{f}}(\mathrm{CH}_{4}) = -76 \mathrm{~kJmol}^{-1}

B) The combustion reaction is


C2H4(g)+3O2(g)→2CO2(g)+2H2O(g)\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g}) + 3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{g}) + 2 \mathrm{H}_{2}\mathrm{O}(\mathrm{g})


On the right hand side of the above expression we know the enthalpy of formation of the CO2\mathrm{CO}_{2} and the H2O\mathrm{H}_{2}\mathrm{O} from their atoms.

It is 2(−394)+2(−286)=−(788+572)=−1360kJ2(-394) + 2(-286) = -(788 + 572) = -1360\mathrm{kJ}. This means that in forming 2 moles of CO2(g)\mathrm{CO}_{2}(\mathrm{g}) and two moles of H2O(l)\mathrm{H}_2\mathrm{O}(\mathrm{l}) from their atoms, 1360kJ1360\mathrm{kJ} are released.

To get there from the left hand side of the equation, we must first form one mole of C2H4\mathrm{C}_2\mathrm{H}_4 from its atoms and then burn it with 3 moles of O2\mathrm{O}_2. ΔHc\Delta \mathrm{H}_{\mathrm{c}} is the heat of combustion.


ΔHf[C2H4(g)]+ΔHc[C2H4(g)]=2ΔH∘f[CO2(g)]+2ΔH∘f[H2O(l)]ΔHf[C2H4(g)]−1411kJ=−1358.6kJ.\begin{array}{l} \Delta \mathrm{Hf} [\mathrm{C}_2\mathrm{H}_4(\mathrm{g})] + \Delta \mathrm{Hc} [\mathrm{C}_2\mathrm{H}_4(\mathrm{g})] = 2 \Delta \mathrm{H}^{\circ} \mathrm{f} [\mathrm{CO}_2(\mathrm{g})] + 2 \Delta \mathrm{H}^{\circ} \mathrm{f} [\mathrm{H}_2\mathrm{O}(\mathrm{l})] \\ \Delta \mathrm{Hf} [\mathrm{C}_2\mathrm{H}_4(\mathrm{g})] - 1411\mathrm{kJ} = -1358.6\mathrm{kJ}. \end{array}

ΔHf[C2H4(g)]=52.4kJ\Delta \mathrm{Hf}[\mathrm{C}_2\mathrm{H}_4(\mathrm{g})] = 52.4\mathrm{kJ}...and since we were talking about one mole of ethylene, that's 52.4 kJ/mole.

Answer: ΔHf(C2H4)=52.4kJmol−1\Delta \mathbf{H}_{\mathrm{f}}(\mathbf{C}_2\mathbf{H}_4) = 52.4\mathrm{kJmol}^{-1}

C) Complete combustion of butane:


2C4H10+13O2→8CO2+10H2O2 \mathrm{C}_4\mathrm{H}_{10} + 13 \mathrm{O}_2 \rightarrow 8 \mathrm{CO}_2 + 10 \mathrm{H}_2\mathrm{O}


Using standard enthalpies of formation, we imagine the reaction takes place by one mole of each starting molecule breaking apart into their elements in their standard states, and then reforming into the products. Obviously this is not really what happens but thermodynamically it is equivalent.

Here's some data:


ΔHf butane(g)=−125kJ/mol4C(s)+5H2(g)→C4H10(g)ΔHfCO2(g)=−394kJ/molC(s)+O2(g)→CO2(g)ΔHfH2O(g)=−286.0kJ/molH2(g)+0.5O2(g)→H2O(l)\begin{array}{l} \Delta \mathrm{Hf} \text{ butane} (\mathrm{g}) = -125 \mathrm{kJ/mol} 4 \mathrm{C}(\mathrm{s}) + 5 \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{C}_4\mathrm{H}_{10}(\mathrm{g}) \\ \Delta \mathrm{Hf} \mathrm{CO}_2(\mathrm{g}) = -394 \mathrm{kJ/mol} \mathrm{C}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) \\ \Delta \mathrm{Hf} \mathrm{H}_2\mathrm{O}(\mathrm{g}) = -286.0 \mathrm{kJ/mol} \mathrm{H}_2(\mathrm{g}) + 0.5 \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2\mathrm{O}(\mathrm{l}) \end{array}


We can simply reverse the sign of the ΔHf\Delta \mathrm{Hf} for the breaking apart of the molecule, as opposed to its formation.

So our reaction comprises:


2C4H10(g)→8C(s)+10H2(g)+125kJ/mol×2=+250.0kJ/mol8O2(g)+8C(s)→8CO2(g)−394kJ/mol×8=−3152kJ/mol10H2(g)+5O2(g)→10H2O(l)−286kJ/mol×10=−2860.0kJ/mol\begin{array}{l} 2 \mathrm{C}_4\mathrm{H}_{10}(\mathrm{g}) \rightarrow 8 \mathrm{C}(\mathrm{s}) + 10 \mathrm{H}_2(\mathrm{g}) + 125 \mathrm{kJ/mol} \times 2 = +250.0 \mathrm{kJ/mol} \\ 8 \mathrm{O}_2(\mathrm{g}) + 8 \mathrm{C}(\mathrm{s}) \rightarrow 8 \mathrm{CO}_2(\mathrm{g}) - 394 \mathrm{kJ/mol} \times 8 = -3152 \mathrm{kJ/mol} \\ 10 \mathrm{H}_2(\mathrm{g}) + 5 \mathrm{O}_2(\mathrm{g}) \rightarrow 10 \mathrm{H}_2\mathrm{O}(\mathrm{l}) - 286 \mathrm{kJ/mol} \times 10 = -2860.0 \mathrm{kJ/mol} \end{array}


Adding these values together:


+250.0kJ/mol+−3152.0kJ/mol+−2860.0kJ/mol=−5762.0kJ/mol\begin{array}{l} +250.0 \mathrm{kJ/mol} + -3152.0 \mathrm{kJ/mol} + -2860.0 \mathrm{kJ/mol} = -5762.0 \mathrm{kJ/mol} \end{array}


We then half this to calculate the standard enthalpy of combustion, since in the above equation we used 2 moles of butane to keep whole numbers:


−5762kJ/mol/2=−2881kJ/mol-5762 \mathrm{kJ/mol} / 2 = -2881 \mathrm{kJ/mol}


Answer: ΔHc(C4H10)=−2881kJmol−1\Delta \mathbf{H}_{\mathrm{c}}(\mathbf{C}_4\mathbf{H}_{10}) = -2881\mathrm{kJmol}^{-1}

D) CH3OH+3/2O2→CO2+2H2O\mathrm{CH}_3\mathrm{OH} + 3/2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} ΔH=−726.4kJ/mol\Delta \mathrm{H} = -726.4\mathrm{kJ/mol}

C(graphite)+O2→CO2ΔH=−393.5kJ/molH2+1/2O2→H2OΔH=−285.8kJ/mol\begin{array}{l} \mathrm{C}(\text{graphite}) + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 \quad \Delta \mathrm{H} = -393.5 \mathrm{kJ/mol} \\ \mathrm{H}_2 + 1/2 \mathrm{O}_2 \rightarrow \mathrm{H}_2\mathrm{O} \quad \Delta \mathrm{H} = -285.8 \mathrm{kJ/mol} \end{array}


Calculate the enthalpy of formation of methanol (CH3OH)(\mathrm{CH}_3\mathrm{OH}) from its elements:


C(graphite)+2H2+1/2O2→CH3OH\mathrm{C}(\text{graphite}) + 2 \mathrm{H}_2 + 1/2 \mathrm{O}_2 \rightarrow \mathrm{CH}_3\mathrm{OH}


As for Hess's law:


−715kJ/mol−1=2x(−286kJ/mol−1)+−394kJ/mol−1−ΔH(CH3OH)ΔH(CH3OH)=−251kJ/mol−1\begin{array}{l} -715 \mathrm{kJ/mol} - 1 = 2 \mathrm{x} (-286 \mathrm{kJ/mol}^{-1}) + -394 \mathrm{kJ/mol}^{-1} - \Delta \mathrm{H}(\mathrm{CH}_3\mathrm{OH}) \\ \Delta \mathrm{H}(\mathrm{CH}_3\mathrm{OH}) = -251 \mathrm{kJ/mol}^{-1} \end{array}


Answer: ΔHf(CH3OH)=−251kJmol−1\Delta \mathbf{H}_{\mathrm{f}}(\mathbf{CH}_3\mathbf{OH}) = -251\mathrm{kJmol}^{-1}

E) The balanced equation I we for ethanol is: C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(g)\mathrm{C}_2\mathrm{H}_5\mathrm{OH}(\mathrm{l}) + 3\mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{CO}_2(\mathrm{g}) + 3\mathrm{H}_2\mathrm{O}(\mathrm{g})

And from that, the three equations we have are:


C+O2→CO2…ΔH=−394kJ/mol\mathrm{C} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 \dots \Delta \mathrm{H} = -394 \mathrm{kJ/mol}

H2+0.5OH→H2O…ΔH=−286kJ/mol\mathrm{H}_2 + 0.5\mathrm{OH} \rightarrow \mathrm{H}_2\mathrm{O} \quad \dots \Delta \mathrm{H} = -286\mathrm{kJ/mol}

2C+3H2+0.5O2→C2H5OH…ΔH=−278kJ/mol2\mathrm{C} + 3\mathrm{H}_{2} + 0.5\mathrm{O}_{2} \rightarrow \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH} \quad \dots \Delta \mathrm{H} = -278\mathrm{kJ/mol}

Then we got:

(278) - (2 x 394) - (3 x 286) = -1368 kJ/mol

Answer: ΔHc\Delta \mathbf{H}_{\mathrm{c}} (C₂H₅OH) = -1368 kJ mol⁻¹

J) C+O2→CO2ΔH1=−394 kJ/mol\mathrm{C} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 \Delta \mathrm{H}_1 = -394 \, \mathrm{kJ/mol}

H2+1/2O2→H2OΔH2=−286 kJ/mol\mathrm{H}_2 + 1/2\mathrm{O}_2 \rightarrow \mathrm{H}_2\mathrm{O} \Delta \mathrm{H}_2 = -286 \, \mathrm{kJ/mol}

CH3COOH+2O2→2CO2+2H2OΔH3=876 kJ/mol\mathrm{CH}_3\mathrm{COOH} + 2\mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \Delta \mathrm{H}_3 = 876 \, \mathrm{kJ/mol}

2C+O2+2H2→CH3COOHΔH4−?2\mathrm{C} + \mathrm{O}_2 + 2\mathrm{H}_2 \rightarrow \mathrm{CH}_3\mathrm{COOH} \Delta \mathrm{H}_4 - ?

Therefore H4=2∗H1+2∗H2−H3=−394−286+876=196 kJ/mol\mathrm{H}_4 = 2^*\mathrm{H}_1 + 2^*\mathrm{H}_2 - \mathrm{H}_3 = -394 - 286 + 876 = 196 \, \mathrm{kJ/mol}

Answer: ΔHf\Delta \mathbf{H}_{\mathrm{f}} (CH₃COOH) = 196 kJ mol⁻¹

F) The balanced equation I we for ethyl acetate is:

CH3COOC2H5(l)+3O2(g)→4CO2(g)+4H2O(g)\mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5(\mathrm{l}) + 3\mathrm{O}_2(\mathrm{g}) \rightarrow 4\mathrm{CO}_2(\mathrm{g}) + 4\mathrm{H}_2\mathrm{O}(\mathrm{g})

And from that, the three equations we have are:

C+O2→CO2ΔH1=−394 kJ/mol\mathrm{C} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 \Delta \mathrm{H}_1 = -394 \, \mathrm{kJ/mol}

H2+1/2O2→H2OΔH2=−286 kJ/mol\mathrm{H}_2 + 1/2\mathrm{O}_2 \rightarrow \mathrm{H}_2\mathrm{O} \Delta \mathrm{H}_2 = -286 \, \mathrm{kJ/mol}

CH3COOC2H5+2O2→4CO2(g)+4H2O(g)ΔH3=−481 kJ/mol\mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5 + 2\mathrm{O}_2 \rightarrow 4\mathrm{CO}_2(\mathrm{g}) + 4\mathrm{H}_2\mathrm{O}(\mathrm{g}) \Delta \mathrm{H}_3 = -481 \, \mathrm{kJ/mol}

CH3COOC2H5→4C+O2+4H2ΔH4−?\mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5 \rightarrow 4\mathrm{C} + \mathrm{O}_2 + 4\mathrm{H}_2 \Delta \mathrm{H}_4 - ?

Therefore H4=2∗H1+2∗H2−H3=(−4.394)+(4.286)+481=−2239 kJ/mol\mathrm{H}_4 = 2^*\mathrm{H}_1 + 2^*\mathrm{H}_2 - \mathrm{H}_3 = (-4.394) + (4.286) + 481 = -2239 \, \mathrm{kJ/mol}

Answer: ΔHc\Delta \mathbf{H}_{\mathrm{c}} (CH₃COOC₂H₅) = -2239 kJ mol⁻¹

B) CH3COOH(l)+C2H5OH(l)→CH3COOC2H5(l)+4H2O(l)ΔH1−?\mathrm{CH}_3\mathrm{COOH}(\mathrm{l}) + \mathrm{C}_2\mathrm{H}_5\mathrm{OH}(\mathrm{l}) \rightarrow \mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5(\mathrm{l}) + 4\mathrm{H}_2\mathrm{O}(\mathrm{l}) \Delta \mathrm{H}_1 - ?

This could be thought of as the result of two hypothetical reactions, going via the elements:

CH3COOH(l)+C2H5OH(l)→4C(s)+5H2(g)+1.5O2(g)ΔH2\mathrm{CH}_3\mathrm{COOH}(\mathrm{l}) + \mathrm{C}_2\mathrm{H}_5\mathrm{OH}(\mathrm{l}) \rightarrow 4\mathrm{C}(\mathrm{s}) + 5\mathrm{H}_2(\mathrm{g}) + 1.5\mathrm{O}_2(\mathrm{g}) \Delta \mathrm{H}_2

followed by 4C(s)+5H2(g)+1.5O2(g)→CH3COOC2H5(l)+4H2OΔH34\mathrm{C}(\mathrm{s}) + 5\mathrm{H}_2(\mathrm{g}) + 1.5\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5(\mathrm{l}) + 4\mathrm{H}_2\mathrm{O} \Delta \mathrm{H}_3

ΔH2\Delta \mathrm{H}_2 is −[ΔHfC2H5OH(l)+ΔHfCH3COOH(l)]-[\Delta \mathrm{H}_{\mathrm{f}} \mathrm{C}_2\mathrm{H}_5\mathrm{OH}(\mathrm{l}) + \Delta \mathrm{H}_{\mathrm{f}} \mathrm{CH}_3\mathrm{COOH}(\mathrm{l})] as it is the reverse of the formation of the compounds from the element, and ΔH3\Delta \mathrm{H}_3 is even more obviously ΔHf(CH3COOC2H5(l))+ΔHf(H2O(l))\Delta \mathrm{H}_{\mathrm{f}}(\mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5(\mathrm{l})) + \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{H}_2\mathrm{O}(\mathrm{l})) as it is the formation of the elements from their compounds. Applying Hess's Law:

ΔH1=ΔH2+ΔH3=−[ΔHfC2H5OH(l)+ΔHfCH3COOH(l)]+[ΔHf(CH3COOC2H5(l))+ΔHf(H2O(l))]\Delta \mathrm{H}_1 = \Delta \mathrm{H}_2 + \Delta \mathrm{H}_3 = -[\Delta \mathrm{H}_{\mathrm{f}} \mathrm{C}_2\mathrm{H}_5\mathrm{OH}(\mathrm{l}) + \Delta \mathrm{H}_{\mathrm{f}} \mathrm{CH}_3\mathrm{COOH}(\mathrm{l})] + [\Delta \mathrm{H}_{\mathrm{f}}(\mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5(\mathrm{l})) + \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{H}_2\mathrm{O}(\mathrm{l}))]

Substituting in appropriate values ΔHf\Delta \mathrm{H}_{\mathrm{f}} (C₂H₅OH (l)) = -1367 kJ/mol; ΔHf\Delta \mathrm{H}_{\mathrm{f}} CH₃COOH (l) = -874 kJ/mol; ΔHf\Delta \mathrm{H}_{\mathrm{f}} (CH₃COOC₂H₅ (l)) = -2239 kJ/mol; ΔHf\Delta \mathrm{H}_{\mathrm{f}} (H₂O (l)) = -286 kJ/mol:

ΔH1=−[(−1367)+(−874)]+[(−2239)+(−286)]=2241−2525=−284 kJ/mol\Delta \mathbf{H}_{1} = -[(-1367) + (-874)] + [(-2239) + (-286)] = 2241 - 2525 = -284 \, \mathrm{kJ/mol}

Answer: -284 kJ/mol


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