Suggest four ways in which the concentration of hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\), could be increased in an equilibrium described by the following equation: \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{4}(g) \quad \Delta H=95 \mathrm{kJ}\)

Short Answer

Expert verified
The concentration of hydrazine can be increased by increasing the pressure, decreasing the temperature, adding more reactants, or using a catalyst.

Step by step solution

01

- Understanding the Equilibrium

Recognize that the given chemical equation is at equilibrium, meaning that the forward and reverse reactions occur at the same rate. The position of equilibrium can be shifted to increase the concentration of a particular reactant or product according to Le Chatelier's Principle.
02

- Applying Le Chatelier's Principle

Le Chatelier's Principle states that if an external change is applied to a system at equilibrium, the system will adjust itself in such a way as to counteract that change. For increasing the concentration of hydrazine, \(\mathrm{N}_{2}\mathrm{H}_{4}\), we can consider changes in concentration, pressure, temperature, and the presence of a catalyst.
03

- Increasing Pressure

Increasing the pressure of the system will shift the equilibrium towards the side with fewer moles of gas. Since the product side has only 1 mole of gas compared to 3 moles of reactants, increasing the pressure will favor the production of hydrazine.
04

- Decreasing Temperature

Since the reaction is endothermic (\(\Delta H=95 \mathrm{kJ}\)), decreasing the temperature would shift the equilibrium to the right in order to absorb heat and increase hydrazine concentration.
05

- Increasing Reactant Concentration

Adding more of either \(\mathrm{N}_{2}\) or \(\mathrm{H}_{2}\) will shift the equilibrium to the right to use up the added reactants, thus increasing the concentration of hydrazine.
06

- Using a Catalyst

Introduction of a catalyst will lower the activation energy and speed up both the forward and reverse reactions equally, resulting in a faster attainment of equilibrium. It will not shift the position of equilibrium but can help reach the equilibrium state where hydrazine concentration is maximized, more rapidly.

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Key Concepts

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

Chemical Equilibrium
In the realm of chemistry, chemical equilibrium is a fascinating state where the rate of the forward reaction equals the rate of the reverse reaction. There are no outward signs of change, yet countless reactions are occurring at the molecular level. This balance, however, is not static and can be altered by external influences.

For instance, imagine a seesaw perfectly balanced in a playground; if you place a weight on one side, the other side will lift off the ground until you adjust the weights to regain balance. In the same way, chemical equilibrium can be disturbed and will adjust accordingly to reach a new state of equilibrium. An understanding of equilibrium is essential to manipulate the concentration of substances in chemical reactions, especially in industrial chemical synthesis and biochemical applications.
Hydrazine Concentration
Speaking of specific substances, hydrazine concentration is crucial in its production processes. Hydrazine is a chemical used as a rocket propellant and in boiler water treatment. To increase its concentration in a reaction, we must shift the equilibrium in favor of its formation.

In our given equation, as the concentration of hydrazine fluctuates, the system seeks to achieve equilibrium by adjusting the rates of the forward and reverse reactions. An increase in raw materials, like nitrogen and hydrogen, could lead to a higher yield of hydrazine. In an industrial setting, efficiency and cost-effectiveness in producing hydrazine are greatly enhanced by understanding how different factors can influence its concentration at equilibrium.
Endothermic Reaction
Digging deeper into the temperature aspects, an endothermic reaction is one that absorbs energy, typically in the form of heat, from its surroundings. The equation presents a classic example: the formation of hydrazine absorbs 95 kJ of heat per mole.

Picture a scenario where you are cold and you put on a jacket to absorb heat from the environment; similarly, in an endothermic reaction, the system 'puts on a jacket' to absorb heat. When the temperature decreases, the reaction swings in favor of the endothermic side to absorb the lost heat, thus increasing the production of hydrazine. This concept is a pivotal part of understanding how temperature control can be used in chemical manufacturing to get the desired product yield.
Equilibrium Shifts
The principle that guides the adjustment of the equilibrium is famously known as Le Chatelier's. This principle implies that an equilibrium shift occurs when an external change is introduced to a system that is in balance. Like nudging a perfectly balanced scale, any change in concentration, pressure, temperature, or catalyst presence can cause the equilibrium to shift.

Assessing our chemical equation, increasing pressure will favor the side with fewer gas molecules, whereas decreasing temperature will favor the endothermic reaction. Both approaches aim to produce more hydrazine by leveraging the natural tendency of the system to counteract changes and maintain a state of equilibrium. Such manipulations of equilibrium conditions are essential techniques in industrial chemistry for optimizing production rates and yields.

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Most popular questions from this chapter

What is the approximate value of the equilibrium constant \(K_{P}\) for the change \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}(l) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}(g)\) at \(25^{\circ} \mathrm{C}\). (The equilibrium vapor pressure for this substance is 570 torr at 25 "C.)

Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.

The following reaction occurs when a burner on a gas stove is lit: \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

The equilibrium constant \(\left(K_{c}\right)\) for this reaction is 5.0 at a given temperature. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) (a) On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of \(\mathrm{CO}, 0.30\) mol of water vapor, and \(0.90 \mathrm{mol}\) of \(\mathrm{H}_{2}\) in a liter. How many moles of \(\mathrm{CO}_{2}\) were there in the equilibrium mixture? (b) Maintaining the same temperature, additional \(\mathrm{H}_{2}\) was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of \(\mathrm{CO}, 0.30\) mol of water vapor, and 1.2 mol of \(\mathrm{H}_{2}\) in a liter. How many moles of \(\mathrm{CO}_{2}\) were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.

Nitrogen and oxygen react at high temperatures. (a) Write the expression for the equilibrium constant \(\left(K_{c}\right)\) for the reversible reaction \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad \Delta H=181 \mathrm{kJ}\) (b) What will happen to the concentrations of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{NO}\) at equilibrium if more \(\mathrm{O}_{2}\) is added? (c) What will happen to the concentrations of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{NO}\) at equilibrium if \(\mathrm{N}_{2}\) is removed? (d) What will happen to the concentrations of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and NO at equilibrium if NO is added? (e) What will happen to the concentrations of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{NO}\) at equilibrium if the volume of the reaction vessel is decreased? (f) What will happen to the concentrations of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{NO}\) at equilibrium if the temperature of the system is increased?

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