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Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$.equilibrium constants \(\mathbf{r}_{\mathbf{p}}\).\(300^{\circ} \mathrm{C}, \quad 4.34 \times 10^{-3}\) \(500^{\circ} \mathrm{C}, \quad 1.45 \times 10^{-5}\) \(600^{\circ} \mathrm{C}, \quad 2.25 \times 10^{-6}\) Is the reaction exothermic or endothermic?

Short Answer

Expert verified
The reaction is exothermic because the equilibrium constant decreases with an increase in temperature, indicating a negative enthalpy change (ΔH).

Step by step solution

01

Observe the equilibrium constants' behavior at different temperatures

Observe the given equilibrium constants (K) and their corresponding temperatures (T): - \(300^{\circ} \mathrm{C}\) and K = \(4.34 \times 10^{-3}\) - \(500^{\circ} \mathrm{C}\) and K = \(1.45 \times 10^{-5}\) - \(600^{\circ} \mathrm{C}\) and K = \(2.25 \times 10^{-6}\) As the temperature increases from \(300^{\circ} \mathrm{C}\) to \(500^{\circ} \mathrm{C}\) and further to \(600^{\circ} \mathrm{C}\), the equilibrium constant decreases.
02

Determine the nature of the reaction

Since the equilibrium constant decreases with an increase in temperature, the reaction is exothermic. The negative enthalpy change (ΔH) causes the decrease in equilibrium constant as the temperature increases. So, the given reaction is exothermic.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
The behavior of chemical reactions can be quite complex, especially when they reach a state called chemical equilibrium. Imagine a busy downtown where traffic is coming from different directions, and at some point, the number of cars entering is equal to the number leaving – that's a simple way of looking at equilibrium in a reaction.

When a reaction reaches chemical equilibrium, the rates of the forward and reverse reactions are equal. This means that the concentrations of the reactants and products remain constant over time, but not necessarily equal, because the system is balanced. It's important to note that equilibrium represents a dynamic balance; even though concentrations remain constant, the reactants are still converting into products and vice versa, but at identical rates.

In the case of the Haber process for producing ammonia, nitrogen and hydrogen gases react to form ammonia, and this reaction can also reverse – ammonia can decompose back into nitrogen and hydrogen. Equilibrium is reached when the rate of formation of ammonia equals the rate of its decomposition.
Equilibrium Constant
The equilibrium constant, represented by the symbol K, is a number that provides a lot of information about a chemical reaction at equilibrium. It's like a scorecard that tells you which team is 'winning' in a sports game where the teams are the reactants and products in a chemical reaction.

Mathematically, it's determined by the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, each raised to the power of their respective coefficients from the balanced chemical equation. For the Haber process the equation looks like this: \( K = \frac{[NH_3]^2}{[N_2][H_2]^3} \).

The magnitude of the equilibrium constant gives us insight into the position of the equilibrium: a larger value, much greater than 1, means there's more product than reactant, while a smaller value, much less than 1, suggests there's more reactant. The given equilibrium constants for the Haber process at different temperatures show this 'score' changes with temperature, implying the position of equilibrium shifts with temperature changes.
Exothermic Reaction
An exothermic reaction is one that releases heat; it's like a natural hand warmer providing toasty warmth on a cold day. In other words, it's a chemical reaction that results in a net release of energy to the surroundings - think of it as the chemical formula for a miniature sun.

The heat released during these reactions can often be felt or measured with a thermometer. In the context of the Haber process, we can infer that it is exothermic because as the temperature is increased, the equilibrium constant decreases. It's like the reaction doesn't 'want' to form as much product at higher temperatures because it's already getting heat from the surroundings, which is enough to satisfy its energy release 'needs'.

Physically, this decrease in the equilibrium constant with increasing temperature supports the idea that heat is a product of the reaction – and according to Le Chatelier's principle, adding more of a product, in this case heat, will shift the equilibrium to favor the reactants, hence the decrease in equilibrium constant values.

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Most popular questions from this chapter

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\). After equilibrium is reached the total pressure is 1.5 atm and \(16 \%\) (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) has dissociated to \(\mathrm{NO}_{2}(g)\) a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C}\) b. If the volume of the cylinder is increased until the total pressure is 1.0 atm (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g)\) c. What percentage (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is dissociated at the new equilibrium position (total pressure \(=\) \(1.00 \mathrm{atm}) ?\)

Suppose \(K=4.5 \times 10^{-3}\) at a certain temperature for the reaction,$$\operatorname{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$.If it is found that the concentration of \(\mathrm{PCl}_{5}\) is twice the concentration of \(\mathrm{PCl}_{3},\) what must be the concentration of \(\mathrm{Cl}_{2}\) under these conditions?

At \(900^{\circ} \mathrm{C}, K_{\mathrm{p}}=1.04\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$.At a low temperature, dry ice (solid \(\mathrm{CO}_{2}\) ), calcium oxide, and calcium carbonate are introduced into a 50.0 -L reaction chamber. The temperature is raised to \(900^{\circ} \mathrm{C},\) resulting in the dry ice converting to gaseous \(\mathrm{CO}_{2}\). For the following mixtures, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium at \(900^{\circ} \mathrm{C} ?\) a. \(655 \mathrm{g} \mathrm{CaCO}_{3}, 95.0 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=2.55 \mathrm{atm}\) b. \(780 \mathrm{g} \mathrm{CaCO}_{3}, 1.00 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{atm}\) c. \(0.14 \mathrm{g} \mathrm{CaCO}_{3}, 5000 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{atm}\) d. \(715 \mathrm{g} \mathrm{CaCO}_{3}, 813 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=0.211 \mathrm{atm}\)

Which of the following statements is(are) true? Correct the false statement(s). a. When a reactant is added to a system at equilibrium at a given temperature, the reaction will shift right to reestablish equilibrium. b. When a product is added to a system at equilibrium at a given temperature, the value of \(K\) for the reaction will increase when equilibrium is reestablished. c. When temperature is increased for a reaction at equilibrium, the value of \(K\) for the reaction will increase. d. When the volume of a reaction container is increased for a system at equilibrium at a given temperature, the reaction will shift left to reestablish equilibrium. e. Addition of a catalyst (a substance that increases the speed of the reaction) has no effect on the equilibrium position.

A \(4.72-\mathrm{g}\) sample of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) was placed in an otherwise empty 1.00 -L flask and heated to \(250 .^{\circ} \mathrm{C}\) to vaporize the methanol. Over time, the methanol vapor decomposed by the \(10=\) following reaction:$$\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)$$.After the system has reached equilibrium, a tiny hole is drilled in the side of the flask allowing gaseous compounds to effuse out of the flask. Measurements of the effusing gas show that it contains 33.0 times as much \(\mathrm{H}_{2}(g)\) as \(\mathrm{CH}_{3} \mathrm{OH}(g) .\) Calculate \(K\) for this reaction at \(250 .^{\circ} \mathrm{C}\).

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