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Chapter 17: Problem 56

Hydrogen cyanide is produced industrially by the following exothermic reaction: $$2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ Is the high temperature needed for thermodynamic or kinetic reasons?

Short Answer

Expert verified
The high temperature needed for hydrogen cyanide production in the given exothermic reaction is primarily for kinetic reasons. Since the reaction is not thermodynamically favored at high temperatures, the high temperature is required to increase the reaction rate, enabling faster production of hydrogen cyanide at an industrial scale.

Step by step solution

01

Understand the effect of temperature on reaction rates

For most chemical reactions, increasing the temperature increases the reaction rate. This is because higher temperatures lead to a higher average kinetic energy of the reacting molecules, which results in more successful collisions between them. The increased reaction rate is generally described by the Arrhenius equation: \(k = Ae^{\frac{-E_a}{RT}}\), where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. Clearly, as the temperature increases, the rate constant k increases, leading to faster reactions.
02

Understand the effect of temperature on equilibrium constants

Temperature also affects equilibrium constants (K) of reactions. For exothermic reactions (\(ΔH < 0\)), increasing the temperature reduces the equilibrium constant, meaning the reaction shifts toward the reactants. On the other hand, for endothermic reactions (\(ΔH > 0\)), increasing the temperature increases the equilibrium constant, favoring the products. This is described by the Van't Hoff equation: \(\frac{d \ln K}{dT} = \frac{ΔH°}{R \cdot T^2}\).
03

Analyze the given reaction

The industrial production of hydrogen cyanide is given by the exothermic reaction: \(2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\). Since this reaction is exothermic, increasing the temperature should result in a lower equilibrium constant and shift the reaction toward the reactants according to thermodynamics. This implies that the high temperature needed for this process is not for thermodynamic reasons.
04

Determine the dominant reason

Since the reaction is not favored by high temperatures for thermodynamic reasons, we can conclude that the high temperature needed for hydrogen cyanide production is primarily for kinetic reasons. High temperature increases the reaction rate, enabling faster production of hydrogen cyanide at an industrial scale.

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Most popular questions from this chapter

You have a 1.00 -L sample of hot water \(\left(90.0^{\circ} \mathrm{C}\right)\) sitting open in a \(25.0^{\circ} \mathrm{C}\) room. Eventually the water cools to \(25.0^{\circ} \mathrm{C}\) while the temperature of the room remains unchanged. Calculate \(\Delta S_{\mathrm{sur}}\) for this process. Assume the density of water is 1.00 $\mathrm{g} / \mathrm{cm}^{3}$ over this temperature range, and the heat capacity of water is constant over this temperature range and equal to 75.4 $\mathrm{J} / \mathrm{K} \cdot$ mol.

Which of the following reactions (or processes) are expected to have a negative value for \(\Delta S^{\circ} ?\) a. $\mathrm{SiF}_{6}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)+\mathrm{SiF}_{4}(g)$ b. $4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)$ c. \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)\) d. $\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)$ e. \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\)

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K)\) . Realizing what $\Delta G^{\circ}$ and K mean, how can you figure out the correct sign?

Predict the sign of \(\Delta S\) for each of the following and explain. a. the evaporation of alcohol b. the freezing of water c. compressing an ideal gas at constant temperature d. dissolving NaCl in water

It is quite common for a solid to change from one structure to another at a temperature below its melting point. For example, sulfur undergoes a phase change from the rhombic crystal structure to the monoclinic crystal form at temperatures above \(95^{\circ} \mathrm{C}\) a. Predict the signs of \(\Delta H\) and \(\Delta S\) for the process $\mathrm{S}_{\text { rhombic}}(s)\longrightarrow \mathrm{S}_{\text { monoclinic }}(s)$ b. Which form of sulfur has the more ordered crystalline structure (has the smaller positional probability)?
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