Gas pipelines are used to deliver natural gas (methane, \(\mathrm{CH}_{4}\) ) to the various regions of the United States. The total volume of natural gas that is delivered is on the order of \(2.7 \times 10^{12} \mathrm{L}\) per day, measured at STP. Calculate the total enthalpy change for combustion of this quantity of methane. (Note: Less than this amount of methane is actually combusted daily. Some of the delivered gas is passed through to other regions.)

Since the volume of methane is given at STP (standard temperature and pressure), we can use the ideal gas law to convert the given volume to moles: \[PV = nRT\] At STP, the temperature (T) is 273.15 K, and the pressure (P) is 1 atm. The ideal gas constant (R) is 0.0821 L atm K^{-1} mol^{-1}. We are given the volume (V) as \(2.7 \times 10^{12} L\) and need to find the number of moles (n). Rearranging the ideal gas law and plugging in the values, we get: \[n = \frac{PV}{RT} = \frac{(1)(2.7 \times 10^{12} L)}{(0.0821)(273.15)} \approx 1.2 \times 10^{11} \text{ mol}\]

The balanced chemical equation for the combustion of methane is: \[\mathrm{CH}_{4} + 2\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}\] From this equation, we can see that for every mole of methane combusted, one mole of carbon dioxide is produced and two moles of water are produced. Since we have \(1.2 \times 10^{11} \text{ mol}\) of methane, we will have the same amount of carbon dioxide and twice the amount of water produced.

We will now calculate the enthalpy change for the combustion of methane using the standard enthalpies of formation (\(\Delta H_f^\circ\)) for the reactants and products. For this, we have: \[\Delta H^\circ = \sum n_\text{products} \Delta H_f^\circ(\text{products}) - \sum n_\text{reactants} \Delta H_f^\circ(\text{reactants})\] The standard enthalpies of formation for the reactants and products are as follows: \[\Delta H_f^\circ(\mathrm{CH}_{4}) = -74.87 \frac{\text{kJ}}{\text{mol}}\] \[\Delta H_f^\circ(\mathrm{O}_{2}) = 0 \frac{\text{kJ}}{\text{mol}}\] \[\Delta H_f^\circ(\mathrm{CO}_{2}) = -393.51 \frac{\text{kJ}}{\text{mol}}\] \[\Delta H_f^\circ(\mathrm{H}_{2}\mathrm{O}) = -241.83 \frac{\text{kJ}}{\text{mol}}\] Applying these values to the equation above, we have: \[\Delta H^\circ = \left[ (1)(-393.51) + (2)(-241.83) \right] - \left[ (1)(-74.87) + (2)(0) \right] = -802.33 \frac{\text{kJ}}{\text{mol}}\]

Finally, we will multiply the enthalpy change for the combustion of one mole of methane by the total moles of methane to obtain the total enthalpy change: \[\Delta H_\text{total} = (1.2 \times 10^{11} \text{mol}) \times (-802.33 \text{ kJ/mol}) = -9.63 \times 10^{13} \text{ kJ}\] So, the total enthalpy change for the combustion of this quantity of methane is approximately \( -9.63 \times 10^{13} \text{ kJ}\).