Question

Consider the combustion of one mole of methane gas:

${\mathrm{CH}}_4\text{(gas)}+2{\mathrm{O}}_2\text{(gas)}⟶{\mathrm{CO}}_2\text{(gas)}+2{\mathrm{H}}_2\mathrm{O}\left(\text{gas}\right)$

The system is at standard temperature $(298\mathrm{K})$and pressure $\left({10}^5\mathrm{Pa}\right)$both before and after the reaction.

(a) First imagine the process of converting a mole of methane into its elemental constituents (graphite and hydrogen gas). Use the data at the back of this book to find $\Delta H$for this process.

(b) Now imagine forming a mole of ${\mathrm{CO}}_2$and two moles of water vapor from their elemental constituents. Determine $\Delta H$for this process.

(c) What is $\Delta H$for the actual reaction in which methane and oxygen form carbon dioxide and water vapor directly? Explain.

(d) How much heat is given off during this reaction, assuming that no "other" forms of work are done?

(e) What is the change in the system's energy during this reaction? How would your answer differ if the${\mathrm{H}}_2\mathrm{O}$ended up as liquid water instead of vapor?

(f) The sun has a mass of$2\times {10}^{30}\mathrm{kg}$and gives off energy at a rate of $3.9\times$${10}^{26}$watts. If the source of the sun's energy were ordinary combustion of a chemical fuel such as methane, about how long could it last?

Short answer

The data at the back of this book to find $∆H$localid="1650341273731" $∆H$for this process. is

localid="1650341277027" $\text{(a)}\mathrm{\Delta }{H}_{C{H}_4}\left(\text{dissociation}\right)=74.81\mathrm{kJ}$

Forming a mole of localid="1650341280022" $∆H$ and two moles of water vapor from their elemental constituents is

localid="1650341283020" $\text{(b)}\mathrm{\Delta }H=-877.15\mathrm{kJ}$

For the actual reaction in which methane and oxygen form carbon dioxide and water vapor directly localid="1650341286125" $\text{(c)}\mathrm{\Delta }H=-802.34\mathrm{kJ}$

Heat is given off during this reaction, assuming that no "other" forms of work are done is

localid="1650341289419" $\text{(d)}Q=-802.34\mathrm{kJ}$

(e) If the water in vapor state localid="1650341292882" $\Delta U=-802.34\mathrm{kJ}$If the water in liquid statelocalid="1650341296979" $\Delta U=-890.36-(-4.95)=-885.41\mathrm{kJ}$

The sun's energy were ordinary combustion of a chemical fuel such as methane, could it last at

localid="1650341300797" $\text{(f)}t=1652\mathrm{y}$

Step-by-step solution

Step1:Temperature Reaction(part a)

(a)Using the enthalpy of the reactants and products, we can calculate how much heat is emitted or absorbed by a chemical reaction. Consider the combustion of methane in the presence of oxygen at a constant temperature of $298°\mathrm{K}$:

${\mathrm{CH}}_4\left(\mathrm{gas}\right)+2{\mathrm{O}}_2\left(\mathrm{gas}\right)\to {\mathrm{CO}}_2\left(\mathrm{gas}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{gas}\right)$

(a) The enthalpy for the formation of methane from elemental carbon (solid) and hydrogen (gas) is calculated using the table at the back of Schroeder's book as follows:

$2{\mathrm{H}}_2\left(\mathrm{gas}\right)+\mathrm{C}\left(\text{solid}\right)\to {\mathrm{CH}}_4\left(\mathrm{gas}\right)$

$2\mathrm{\Delta }{H}_{H_2}+\mathrm{\Delta }{H}_C\to \mathrm{\Delta }{H}_{C{H}_4}$

$2\left(0\right)+\left(0\right)\to -74.81$

Step2:determine∆H(part b)

Similarly, the enthalpy for producing one mole of carbon dioxide from elemental carbon (solid) and oxygen (gas) is$O_2\left(gas\right)+C\left(\text{solid}\right)\to C_2\left(\text{gas}\right)$

localid="1650341326563" $\begin{array}{c}\mathrm{\Delta }{H}_{O_2}+\mathrm{\Delta }{H}_C\to \mathrm{\Delta }{H}_{C{O}_2}\\ \left(0\right)+\left(0\right)\to -393.51\end{array}$

Solocalid="1650341329756" $\Delta H$for the formation of methane is therefore:

localid="1650341332718" $\mathrm{\Delta }{H}_{C{O}_2}-\mathrm{\Delta }{H}_C-\mathrm{\Delta }{H}_{O_2}=-393.51-0-0=-393.51\mathrm{kJ}$

localid="1650341335575" $\to {\mathrm{\Delta H}}_{{\mathrm{CO}}_2}\left(\text{formation}\right)=-393.51\mathrm{kJ}$

The enthalpy of forming two moles of vapor from elemental oxygen (gas) and hydrogen (gas) is as follows

localid="1650341339663" $2{\mathrm{H}}_2\left(\mathrm{gas}\right)+\mathrm{O}\left(\mathrm{gas}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{gas}\right)$

localid="1650341342662" $\begin{array}{c}2\mathrm{\Delta }{H}_{H_2}+\mathrm{\Delta }{H}_O\to 2\mathrm{\Delta }{H}_{H_{2}O}\\ 2\left(0\right)+\left(0\right)\to 2(-241.82)\end{array}$

Solocalid="1650341346216" $\Delta H$As a result, the formation of two moles of water is required.:

localid="1650341349932" $2\mathrm{\Delta }{H}_{H_{2}O}-\mathrm{\Delta }{H}_O-\mathrm{\Delta }{H}_{H_2}=2(-241.82)-0-0=-483.64\mathrm{kJ}$

localid="1650341353807" $\to {\mathrm{\Delta H}}_{2{\mathrm{H}}_2\mathrm{O}}\left(\text{formation}\right)=-483.64\mathrm{kJ}$

The enthalpy of producing one mole of CO2 and two moles of water is:

localid="1650341357759" $\left.\mathrm{\Delta }H=\mathrm{\Delta }{H}_{2H_{2}O}\text{(formation}\right)+\mathrm{\Delta }{H}_{C{O}_2}\left(\text{formation}\right)=-483.64+-393.51$

localid="1650341361917" $\to \mathrm{\Delta }H=-877.15\mathrm{kJ}$

Step3:Describe which methane and oxygen(part c)

(c)$\Delta H$for the reaction as a whole is;$\mathrm{\Delta }H=\mathrm{\Delta }{H}_{\text{prod}}-\mathrm{\Delta }{H}_{\text{react}}$

where localid="1650341370451" $\Delta H_{\text{prod}}$is the enthalpy of producing one mole of CO2 and two moles of water. (calculated in (b)), and localid="1650341373459" $\Delta H_{\text{react}}$is the enthalpy of methane dissociation (calculated in localid="1650341376617" $\mathbf{(}\mathbf{a})$), It is worth noting that the enthalpy of oxygen dissociation is zero. so:

localid="1650341379957" $\mathrm{\Delta }H=\mathrm{\Delta }{H}_{C_2}\left(\text{formation}\right)+\mathrm{\Delta }{H}_{2H_{2}O}\left(\text{formation}\right)-\mathrm{\Delta }{H}_{C{H}_4}\left(\text{dissociation}\right)$

localid="1650341383103" $\to \mathrm{\Delta }H=-393.51-483.64+74.81=-802.34\mathrm{kJ}$

Step4:how much heat is produced during this reaction (part d)

(d) The enthalpy at constant pressure is given by:

$\mathrm{\Delta }H=Q+W_{other}$

Because no other work is being done on the reaction, the amount of heat is therefore:

$\to Q=\mathrm{\Delta }H=-802.34\mathrm{kJ}$

Step5:During this reaction, the system energy(part e)

(e) If all four compounds in the main equation are gases and the temperature is the same on both sides, there will be no volume change$∆V=0$because three moles of gas are present both before and after the reaction As a result, for one mole of methane, the entire change in enthalpy is due to a change in internal energy U.

$\begin{array}{c}\mathrm{\Delta }H=\mathrm{\Delta }U+P\mathrm{\Delta }V\\ \mathrm{\Delta }U=\mathrm{\Delta }H=-802.34\mathrm{kJ}\end{array}$

If water is produced as a liquid rather than a vapor, the enthalpy for producing two moles of liquid water from elemental oxygen (gas) and hydrogen (gas) is:

$2{\mathrm{H}}_2\left(\mathrm{gas}\right)+\mathrm{O}\left(\mathrm{gas}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\text{liquid}\right)$

$\begin{array}{c}2\mathrm{\Delta }{H}_{H_2}+\mathrm{\Delta }{H}_O\to 2\mathrm{\Delta }{H}_{H_{2}O}\\ 2\left(0\right)+\left(0\right)\to 2(-285.83)\end{array}$

So $\Delta H$as a result of the formation of methane:

$2\mathrm{\Delta }{H}_{H_{2}O}-\mathrm{\Delta }{H}_O-\mathrm{\Delta }{H}_{H_2}=2(-285.83)-0-0=-571.66\mathrm{kJ}$

So, $\Delta H$for the reaction as a whole is:

$\to \mathrm{\Delta }H=-393.51-571.66+74.81=-890.36\mathrm{kJ}$

This time, the final volume is $\frac{1}{3}$of the initial volume, Because the two moles of water have condensed into a liquid with a negligible volume in comparison to the gases. As a result, the environment is effective:

$W=P\mathrm{\Delta }V=RT\mathrm{\Delta }n=RT\left(n_f-n_i\right)$

As a result, the change in internal energy is discovered from:

$\mathrm{\Delta }U=\mathrm{\Delta }H-W$

$\to \mathrm{\Delta }U=-890.36-(-4.95)=-885.41\mathrm{kJ}$

The latent heat of vaporization should be the difference between this value and when the water is produced as vapor at ${}^{\circ }298\mathrm{K}.$

Step6:The sun been in existence (part f)

(f) Suppose the Sun with a mass of around $2\times {10}^{33}\mathrm{g}$and luminosity of$3.839\times {10}^{26}$watts Its energy source was the combustion of methane and oxygen. The molar weights of methane and molecular oxygen are approximately$16mg$and $32mg$, respectively. The chemical reaction of methane consumption is as follows:

${\mathrm{CH}}_4\left(\mathrm{gas}\right)+2{\mathrm{O}}_2\left(\mathrm{gas}\right)\to {\mathrm{CO}}_2\left(\mathrm{gas}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{gas}\right)$

So, if the Sun is made up of one part methane and two parts oxygen, the mole ratio is:

$\frac{n_{C{H}_4}}{n_{O_2}}=\frac{1}{2}$

The mass ratio is then:

$\frac{m_{C{H}_4}}{m_{O_2}}=\frac{\text{mass of one mole of methane}\times n_{C{H}_4}}{\text{mass of one mole of oxygen}\times n_{O_2}}$

$\frac{m_{C{H}_4}}{m_{O_2}}=\frac{16\times 1}{32\times 2}=\frac{1}{4}$

As a result, the mass of methane in the Sun is:

$m_{C{H}_4}=\text{ratio of methane mass}\times \text{Sun mass}$

$m_{C{H}_4}=\frac{1}{4+1}\times 2\times {10}^{33}=4\times {10}^{32}\mathrm{g}$

As a result, the number of moles of methane in the Sun is:

$n_{C{H}_4}=\frac{\text{methane mass in the sun}}{\text{mass of one mole of methane}}$

$n_{C{H}_4}=\frac{4\times {10}^{32}}{16}=2.5\times {10}^{31}\mathrm{mol}$

Step7:How long can the sun last if it has a mass?(part f)

(f)Assuming that the water is produced as vapor, the Sun could generate a total energy of:

$E=\text{number of methane moles}\times \text{energy of one mole consumption}$

where the energy of one mole consumption is substituted$802.34\mathrm{kJ}$, so:

$E=2.5\times {10}^{31}\times 802.34=2\times {10}^{34}\mathrm{kJ}$

so it would burn out after a time interval of:

$\text{Power}=\frac{E}{t}\to t=\frac{E}{\text{Power}}$

substitute, where Power$=3.839\times {10}^{26}$watts

$$\begin{gathered} t=\frac{2\times {10}^{34}\mathrm{kJ}}{3.839\times {10}^{26}\mathrm{J}\cdot {\mathrm{s}}^{-1}} \\ =\frac{2\times {10}^{37}\mathrm{J}}{3.839\times {10}^{26}\mathrm{J}\cdot {\mathrm{s}}^{-1}} \\ =5.2\times {10}^{10}\mathrm{s} \end{gathered}$$

$t=1652\mathrm{y}$

We're pretty sure the Sun's power source isn't chemical reactions!