About 75 percent of hydrogen for industrial use is produced by the steam- reforming process. This process is carried out in two stages called primary and secondary reforming. In the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at \(800^{\circ} \mathrm{C}\) to give hydrogen and carbon monoxide: $$\begin{array}{r}\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \\\\\Delta H^{\circ}=260 \mathrm{~kJ} /\mathrm{mol}\end{array}$$ The secondary stage is carried out at about \(1000^{\circ} \mathrm{C}\), in the presence of air, to convert the remaining methane to hydrogen: $$\begin{array}{r}\mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \\\\\Delta H^{\circ}=35.7 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stage? (b) The equilibrium constant \(K_{\mathrm{c}}\) for the primary stage is 18 at \(800^{\circ} \mathrm{C}\). (i) Calculate \(K_{P}\) for the reaction. (ii) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium?

Step by step solution

01

Analyzing Conditions for Product Formation

To predict the effect of temperature change, consideration is given to whether the reaction is exothermic or endothermic. In both the primary and secondary stages, the reactions are endothermic as suggested by positive ∆H values. Therefore, an increase in temperature will favor product formation as per Le Châtelier’s principle. Similarly, pressure changes will affect gas-phase reactions where there is a change in the total number of moles. In the primary stage, there is a decrease in moles from left to right, so a higher pressure would shift the equilibrium towards the reactants. Conversely, the secondary stage has balanced moles on both sides, so changes in pressure would not have an effect.

02

Calculating \(K_{P}\)

The relationship between \(K_{C}\) and \(K_{P}\) is given by the equation \(K_{P} = K_{C}(RT)^{\Delta{n}}\), where \(R\) is the universal gas constant (0.0821 L·atm/(K·mol)), \(\Delta{n}\) is the change in moles of gaseous reactants and products and \(T\) is the temperature in Kelvin. In the primary stage, \(\Delta{n} = 4 - 2 = 2\), and \(T = 800^{\circ} C = 1073K\). Therefore, \(K_{P} = 18 * (0.0821*1073)^2 = 276KPa\)

03

Determining Pressures at Equilibrium

If the pressures of methane and steam were both 15 atm initially, let the amount that reacts be \(x\). Then at equilibrium, the pressures are \(15-x\) for \(CH_{4}(g)\) and \(H_{2}O(g)\), and coincide with the formation of \(x\) moles of \(CO(g)\) and \(3x\) moles of \(H_{2}(g)\). Substituting into the equation for \(K_{P}\), we get \(K_{P} = [x][3x]^3 / [(15-x)]^2 = 276KPa\). Solving for \(x\) yields pressures at equilibrium; \( CH_{4} = H_{2}O = 15 - 1.62 = 13.38 atm\), \( CO = 1.62 atm\), \( H_{2} = 4.86 atm\)

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