91Ó°ÊÓ

(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 $298°\mathrm{K}$:

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

(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:

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

$2\mathrm{Δ}{H}_{H_2}+\mathrm{Δ}{H}_C→\mathrm{Δ}{H}_{C{H}_4}$

$2\left(0\right)+\left(0\right)→−74.81$

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

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

Solocalid="1650341329756" $\mathrm{Δ}H$for the formation of methane is therefore:

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

localid="1650341335575" $→{\mathrm{Δ\pm á}}_{{\mathrm{CO}}_2}\left(\text{formation}\right)=−393.51\mathrm{kJ}$

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

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

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

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

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

localid="1650341353807" $→{\mathrm{Δ\pm á}}_{2{\mathrm{H}}_2\mathrm{O}}\left(\text{formation}\right)=−483.64\mathrm{kJ}$

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

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

localid="1650341361917" $→\mathrm{Δ}H=−877.15\mathrm{kJ}$

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

where localid="1650341370451" $\mathrm{Δ}{H}_{\text{prod}}$is the enthalpy of producing one mole of CO2 and two moles of water. (calculated in (b)), and localid="1650341373459" $\mathrm{Δ}{H}_{\text{react}}$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" $\mathrm{Δ}H=\mathrm{Δ}{H}_{C_2}\left(\text{formation}\right)+\mathrm{Δ}{H}_{2H_{2}O}\left(\text{formation}\right)−\mathrm{Δ}{H}_{C{H}_4}\left(\text{dissociation}\right)$

localid="1650341383103" $→\mathrm{Δ}H=−393.51−483.64+74.81=−802.34\mathrm{kJ}$

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

$\mathrm{Δ}H=Q+W_{other}$

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

$→Q=\mathrm{Δ}H=−802.34\mathrm{kJ}$

(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.

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

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:

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

$\begin{array}{c}2\mathrm{Δ}{H}_{H_2}+\mathrm{Δ}{H}_O→2\mathrm{Δ}{H}_{H_{2}O}\\ 2\left(0\right)+\left(0\right)→2(−285.83)\end{array}$

So $\mathrm{Δ}H$as a result of the formation of methane:

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

So, $\mathrm{Δ}H$for the reaction as a whole is:

$→\mathrm{Δ}H=−393.51−571.66+74.81=−890.36\mathrm{kJ}$

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:

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

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

$\mathrm{Δ}U=\mathrm{Δ}H−W$

$→\mathrm{Δ}U=−890.36−(−4.95)=−885.41\mathrm{kJ}$

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

(f) Suppose the Sun with a mass of around $2×{10}^{33}\mathrm{g}$and luminosity of$3.839×{10}^{26}$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:

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

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:

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

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

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

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

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

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

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

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

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

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

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

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

so it would burn out after a time interval of:

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

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

$$\begin{gathered} t=\frac{2×{10}^{34}\mathrm{kJ}}{3.839×{10}^{26}\mathrm{J}â‹\dots {\mathrm{s}}^{−1}} \\ =\frac{2×{10}^{37}\mathrm{J}}{3.839×{10}^{26}\mathrm{J}â‹\dots {\mathrm{s}}^{−1}} \\ =5.2×{10}^{10}\mathrm{s} \end{gathered}$$

$t=1652\mathrm{y}$

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