The conversion of natural gas, which is mostly methane, into...

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$, is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ In practice, this reaction is carried out in the presence of oxygen: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Using the data in Appendix $C,$ calculate $K$ for these reactions at $25^{\circ} \mathrm{C}$ and $500{ }^{\circ} \mathrm{C}$. (b) Is the difference in $\Delta G^{\circ}$ for the two reactions due primarily to the enthalpy term $(\Delta H)$ or the entropy term $(-T \Delta S) ?(\mathbf{c})$ Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7 . (d) The reaction of $\mathrm{CH}_{4}$ and $\mathrm{O}_{2}$ to form $\mathrm{C}_{2} \mathrm{H}_{6}$ and $\mathrm{H}_{2} \mathrm{O}$ must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$, is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen:
$$
2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g)
$$
In practice, this reaction is carried out in the presence of
oxygen:
$$
2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g)
$$
(a) Using the data in Appendix $C,$ calculate $K$ for these reactions at $25^{\circ} \mathrm{C}$ and $500{ }^{\circ} \mathrm{C}$. (b) Is the difference in $\Delta G^{\circ}$ for the two reactions due primarily to the enthalpy term $(\Delta H)$ or the entropy term $(-T \Delta S) ?(\mathbf{c})$ Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7 . (d) The reaction of $\mathrm{CH}_{4}$ and $\mathrm{O}_{2}$ to form $\mathrm{C}_{2} \mathrm{H}_{6}$ and $\mathrm{H}_{2} \mathrm{O}$ must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

Key Concepts

Equilibrium Constant and Gibbs Free Energy

This concept involves the intrinsic relationship between the equilibrium constant (K) and the standard Gibbs free energy change (?G°) of a reaction, commonly expressed by the formula ?G° = -RT ln K. It highlights how the position of equilibrium changes with temperature, and it allows one to predict the spontaneity of a reaction based on computed or experimental K values, which are vital for understanding reaction feasibility under different conditions.

Enthalpy and Entropy Contributions to Reaction Spontaneity

This concept focuses on how the overall free energy change for a reaction is derived from both the enthalpy change (?H) and entropy change (?S), captured in the equation ?G = ?H - T?S. It emphasizes that the temperature-dependent entropy term can significantly affect the spontaneity of reactions, with the balance between heat release/absorption and disorder change determining whether a process proceeds spontaneously.

Coupled Reactions and Driving Nonspontaneous Processes

This concept explains how a nonspontaneous reaction can be driven forward by coupling it with another reaction that is spontaneous. In industrial and biochemical applications, reactions that do not occur on their own can proceed when linked to exergonic processes, enabling the overall reaction sequence to be feasible through energy transfer mechanisms.

Competing Reactions and Reaction Selectivity

This concept refers to the challenge of avoiding undesirable side reactions that may occur under similar conditions to the target reaction. For instance, in systems where methane and oxygen are present, careful control is required to prevent complete combustion (forming CO2 and water), ensuring that the desired pathway, such as conversion to higher hydrocarbons, is favored over competing, less useful reactions.

Transcript

00:01 Alright, so here is the problem 100 in chapter 19.

00:05 Part a, using the data in appendix c, calculate the k value for this reaction is at 25 and 500 degrees.

00:14 So the way we calculate for k value is using this formula, delta g0, equal to negative rt long k.

00:24 And at the same time, delta g0 is equal to delta h0 minus t delta s0.

00:33 So we use the second formula to get the delta g value at two different temperatures.

00:39 Then the values of delta h0 and delta s0 can be calculated using the data from the appendix.

00:46 So for the first reaction, that is the 2 -ch4 gas giving c2h6.

00:59 Gas and h2 gas.

01:03 For this reaction, we have delta h0 equal to the delta h0 of c2h6 gas plus the delta h0 of hydrogen gas minus two times the delta h0 of the ch4 gas.

01:28 And this is equal to negative 804.

01:32 0 .68 plus 0 minus 2 times negative 74 .8.

01:40 This is equal to 64 .92 kilojoules.

01:47 Then we find for the delta s0 value of the reaction that is equal to the s0 of c2h6 gas plus s0 of hydrogen gas minus 2 times s0 of ch4.

02:04 Gas this will be equal to 229 .5 plus one 30 .58 minus 2 times 186 .3 this is equal to negative 12 .52 joules per kelvin and then we want to convert this into kilojoules so this will be equal to negative 12.

02:38 0 .52 times 10 to minus 3 kilojoules per kelvin.

02:46 So that's the delta h and delta s for the first reaction.

02:52 So now at 25 degree, delta g will be equal to delta g.

03:08 Delta g will be equal to delta h minus t delta s.

03:14 So this will be equal to 46 .92 kilojoules minus 25 plus 273 kelvin times delta s is negative 12 .52 times 10 to minus 3 kilojoules per kelvin.

03:44 So this will get a number of 68 .7 kilojoules.

03:52 So if delta g equal to this value, we have delta g equal to negative rt long k.

04:02 So long k will be equal to negative delta g divided by rt, that is equal to negative 6 .8 .7 kilojoules divided by 8 .34 times 10 to minus 3, kilojoules per kelvin times 2 .3 ,000.

04:23 298 kelvin and this will give the number of negative 27 .7.

04:31 So the value of k is equal to e to the negative 27 .7 and that is approximately 9 .25 times 10 to the minus 13.

04:53 So that's the case at 25 degree.

04:57 So now at 500 degree, for this case, case, the delta g will be equal to 64 .9 kilojoules minus 500 plus 273 kelvin times the delta axis negative 12 .52 times 10 to minus 3 kilojoules per kelvin.

05:36 And this will give you a number of 74 .6 kilojoules.

05:43 The long k equal to negative delta g divided by r t that is equal to negative 74 .6 kilojoules divided by 8 .314 times 10 to minus 3 kilojoules per kelvin times 773 kelvin and this will be equal to negative 11 so the k value will be equal to 9 .1 times 10 to minus 6.

06:25 All right, so this is the k value at 500 degree, and this is the k value at 25 degree, both for the first reaction.

06:45 So now for the second reaction, that is the 2c4 gas plus one half of oxygen given c2h6 gas and water gas for this reaction we are also going to calculate the delta h0 and the delta s zero so delta h0 for this reaction will be the delta h0 of c2 h6 gas plus delta h0 of of h2o gas minus 2 times delta h value of ch4 gas minus one half of the delta h value of oxygen gas.

07:46 So this is equal to negative 84 .68 plus negative 241 .82 minus 2 times negative 74 .8 minus 1 half times 0, and you will get a number of negative 176 .9 kilojoules.

08:14 For delta s0, you're going to use s0 of c2h6 gas plus s0 of water gas minus 2 times, s -0 of ch4 gas minus 1 half as 0 of the oxygen gas.

08:40 So this is equal to 2 to 9 .5 plus 188 .83 minus 2 times 186 .3 minus 1⁄2 .186 .3 minus 1⁄25 of 205 .5.

09:03 This is equal to negative 56 .77 joules per kelvin.

09:11 So converting that to kilojoules, we will have negative 56 .77 times 10 to minus 3 kilojoules per kelvin...

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