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Suppose that you worked at the U.S. Patent Office and a patent application came across your desk claiming that a newly developed catalyst was much superior to the Haber catalyst for ammonia synthesis because the catalyst led to much greater equilibrium conversion of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) into \(\mathrm{NH}_{3}\) than the Haber catalyst under the same conditions. What would be your response?

Short Answer

Expert verified
The claim that the newly developed catalyst has a much superior performance than the Haber catalyst due to its greater equilibrium conversion of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) into \(\mathrm{NH}_{3}\) is not valid. Catalysts can only affect the rate of reaction, not the equilibrium constant. Therefore, the proposed improved conversion cannot be attributed to the catalyst itself. To support the claim that the new catalyst is superior, additional evidence is required, such as a faster reaction rate or improved efficiency in real-world application conditions.

Step by step solution

01

Understanding chemical equilibrium

Chemical equilibrium is a state in which the rate of the forward reaction is equal to the rate of the reverse reaction. It's important to note that when a reaction reaches its equilibrium point, the concentrations of reactants and products remain constant. This is described by the equilibrium constant (K), which is the ratio of the concentrations of products to reactants raised to their respective stoichiometric coefficients. The equilibrium constant is determined solely by the reaction conditions, such as temperature and pressure, and not by catalysts. Catalysts generally increase the rate of a reaction by offering an alternative pathway with lower activation energy but do not affect the equilibrium concentrations of reactants and products.
02

Discussing the claim

The claim made by the patent application states that the newly developed catalyst leads to a much greater equilibrium conversion of N2 and H2 into NH3 than the Haber catalyst under the same conditions. However, as discussed earlier, catalysts can increase the rate of reaction, but they do not alter the equilibrium concentrations of reactants and products. Therefore, the claim that the new catalyst affects the equilibrium conversion is not valid.
03

Responding to the claim

Based on the understanding of chemical equilibrium and the role of catalysts, we can respond to the claim as follows: "The claim that the newly developed catalyst has a much superior performance than the Haber catalyst due to its greater equilibrium conversion of N2 and H2 into NH3 is not valid. Catalysts can only affect the rate of reaction, not the equilibrium constant. Therefore, the proposed improved conversion cannot be attributed to the catalyst itself. To support the claim that the new catalyst is superior, additional evidence is required, such as a faster reaction rate or improved efficiency in real-world application conditions."

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\) the reaction $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ has \(K_{p}=0.120\). A 5.00-L flask is charged with \(0.300 \mathrm{~g}\) of pure \(\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) at \(25^{\circ} \mathrm{C}\). Solid \(\mathrm{NH}_{4} \mathrm{HS}\) is then added until there is excess unreacted solid remaining. (a) What is the initial pressure of \(\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) in the flask? (b) Why does no reaction occur until \(\mathrm{NH}_{4} \mathrm{HS}\) is added? (c) What are the partial pressures of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) at equilibrium? (d) What is the mole fraction of \(\mathrm{H}_{2} \mathrm{~S}\) in the gas mixture at equilibrium? (e) What is the minimum mass, in grams, of \(\mathrm{NH}_{4} \mathrm{HS}\) that must be added to the flask to achieve equilibrium?

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.16), \(K_{p}\) for the reaction $$ 2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) $$ is about \(1 \times 10^{-13}\). Assuming that the exhaust gas (total pressure \(1 \mathrm{~atm}\) ) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2}\), and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the above reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the reaction above?

An equilibrium mixture of \(\mathrm{H}_{2}, \mathrm{I}_{2}\), and \(\mathrm{HI}\) at \(458^{\circ} \mathrm{C}\) contains \(0.112 \mathrm{~mol} \mathrm{H}_{2}, 0.112 \mathrm{~mol} \mathrm{I}_{2}\), and \(0.775 \mathrm{~mol} \mathrm{HI}\) in a 5.00-L vessel. What are the equilibrium partial pressures when equilibrium is reestablished following the addition of \(0.100 \mathrm{~mol}\) of \(\mathrm{HI}\) ?

A mixture of \(\mathrm{H}_{2} \mathrm{~S}\), and \(\mathrm{H}_{2} \mathrm{~S}\) is held in a 1.0-L vessel at \(90^{\circ} \mathrm{C}\) until the following equilibrium is achieved: $$ \mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g) $$ At equilibrium the mixture contains \(0.46 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(0.40 \mathrm{~g} \mathrm{H}_{2}\). (a) Write the equilibrium- constant expression for this reaction. (b) What is the value of \(K_{c}\) for the reaction at this temperature? (c) Why can we ignore the amount of \(S\) when doing the calculation in part \((b)\) ?

The following equilibria were attained at \(823 \mathrm{~K}\) : $$ \begin{array}{ll} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) & K_{c}=67 \\ \mathrm{CoO}(s)+\mathrm{CO}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g) & K_{c}=490 \end{array} $$ Based on these equilibria, calculate the equilibrium constant for \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) at \(823 \mathrm{~K}\)

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