Substitute natural gas (SNG) is a gaseous mixture containing \(\mathrm{CH}_{4}(\mathrm{g})\) that can be used as a fuel. One reaction for the production of SNG is $$\begin{aligned} 4 \mathrm{CO}(\mathrm{g})+8 \mathrm{H}_{2}(\mathrm{g}) & \longrightarrow \\ 3 \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & \Delta H^{\circ}=? \end{aligned}$$ Use appropriate data from the following list to determine \(\Delta H^{\circ}\) for this SNG reaction. $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ \quad\quad\quad\quad\quad\quad\quad\quad\qquad \Delta H^{\circ}=-110.5 \mathrm{k} \mathrm{J} \end{array}$$$$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-283.0 \mathrm{kJ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\begin{array}{l} \text { C(graphite) }+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g}) \\ \qquad \Delta H^{\circ}=-74.81 \mathrm{kJ} \end{array}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+& 2 \mathrm{H}_{2} \mathrm{O}(1) \\ & \Delta H^{\circ}=-890.3 \mathrm{kJ} \end{aligned}$$

Short Answer

Expert verified
The enthalpy change \(\Delta H^∘\) for the production of SNG is \(-2699.93 kJ\)

Step by step solution

01

Identify the reactions that make up the SNG production reaction

The production of SNG is given by the reaction:\[4 CO(g) + 8 H2(g) → 3 CH4(g) + CO2(g) + 2 H2O(l)\]This can be broken down into the following reactions from the given list: \[4 [C(graphite) + 0.5 O2(g) → CO(g)]\]\[4 [2 H2(g) + O2(g) → 2 H2O(l) ]\]\[3 [C(graphite) + 2 H2(g) → CH4(g)]\]and\[1 [CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)]\]
02

Calculate the enthalpy changes of the individual reactions

\[ΔH°_1 = 4 × (-110.5 kJ) = -442 kJ\] for the first reaction.\[ΔH°_2 = 4 × (-285.8 kJ) = -1143.2 kJ\] for the second reaction.\[ΔH°_3 = 3 × (-74.81 kJ) = -224.43 kJ\] for the third reaction.\[ΔH°_4 = 1 × (-890.3 kJ) = -890.3 kJ\] for the final reaction.
03

Calculate the total enthalpy change of the reaction

The total enthalpy change for the reaction will be the sum of the enthalpy changes of the individual steps. This can be calculated as \[ ΔH^∘ = ΔH°_1 + ΔH°_2 + ΔH°_3 + ΔH°_4 = -442 kJ + -1143.2 kJ + -224.43 kJ + -890.3 kJ = -2699.93 kJ \]

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Most popular questions from this chapter

Refer to Example \(7-3 .\) Based on the heat of combustion of sucrose established in the example, what should be the temperature change \((\Delta T)\) produced by the combustion of \(1.227 \mathrm{g} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) in a bomb calorimeter assembly with a heat capacity of \(3.87 \mathrm{kJ} /^{\circ} \mathrm{C} ?\)

Care must be taken in preparing solutions of solutes that liberate heat on dissolving. The heat of solution of \(\mathrm{NaOH}\) is \(-44.5 \mathrm{kJ} / \mathrm{mol} \mathrm{NaOH} .\) To what maximum temperature may a sample of water, originally at \(21^{\circ} \mathrm{C},\) be raised in the preparation of \(500 \mathrm{mL}\) of \(7.0 \mathrm{M}\) NaOH? Assume the solution has a density of \(1.08 \mathrm{g} / \mathrm{mL}\) and specific heat of \(4.00 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

Given the following information: $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{NH}_{3}(\mathrm{g})\quad\quad\quad\quad\Delta H_{1}^{\circ}$$ $$\mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H_{2}^{\circ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\quad\quad\quad\Delta H_{3}^{\circ}$$ Determine \(\Delta H^{\circ}\) for the following reaction, expressed in terms of \(\Delta H_{1}^{\circ}, \Delta H_{2}^{\circ},\) and \(\Delta H_{3}^{\circ}\) $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g}) \quad \Delta H^{\circ}=?$$

A 1.22 kg piece of iron at \(126.5^{\circ} \mathrm{C}\) is dropped into \(981 \mathrm{g}\) water at \(22.1^{\circ} \mathrm{C} .\) The temperature rises to \(34.4^{\circ} \mathrm{C} .\) What will be the final temperature if this same piece of iron at \(99.8^{\circ} \mathrm{C}\) is dropped into \(325 \mathrm{mL}\) of glycerol, \(\mathrm{HOCH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}(1)\) at \(26.2^{\circ} \mathrm{C} ?\) For glycerol, \(d=1.26 \mathrm{g} / \mathrm{mL} ; C_{n}=219 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\).

Is it possible for a chemical reaction to have \(\Delta U<0\) and \(\Delta H>0 ?\) Explain.

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