In a flask, colourless \(\mathrm{N}_{2} \mathrm{O}_{4}\) is in equilibrium with brown coloured \(\mathrm{NO}_{2}\). At equilibrium, when the flask is heated at \(100^{\circ} \mathrm{C}\) the brown colour deepens and on cooling it becomes less coloured. The change in enthalpy, \(\Delta H\), for the system is (a) negative (b) positive (c) zero (d) undefined

Short Answer

Expert verified
The change in enthalpy, \(\Delta H\), for the system is positive.

Step by step solution

01

Understanding Le Chatelier's Principle

Le Chatelier's Principle indicates that if an external change is applied to a system at equilibrium, the system adjusts itself to partially counteract the change. In this case, heating causes the equilibrium to shift in a direction that absorbs heat.
02

Analyzing the Effect of Temperature on the Equilibrium

Since heating the system deepens the brown color, which is a characteristic of \(\mathrm{NO}_{2}\), we can infer that the equilibrium shifts towards the production of more \(\mathrm{NO}_{2}\). This is an endothermic reaction because it requires the absorption of heat.
03

Determining the Sign of the Change in Enthalpy \(\Delta H\)

For an endothermic process, the change in enthalpy is positive because the system absorbs energy from the surroundings. Therefore, the reaction \(\mathrm{N}_{2} \mathrm{O}_{4} \rightarrow 2 \mathrm{NO}_{2}\) becomes more favorable at higher temperatures, indicating a positive \(\Delta H\).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentrations of reactants and products remain constant over time, although both reactions are still occurring. This balance can be represented by a reversible arrow in chemical equations (
\( A + B \rightleftharpoons C + D \)
).

Le Chatelier's Principle plays a crucial role in understanding how this balance can be disturbed. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. In the context of our exercise, when temperature is increased, the equilibrium system of colourless \(\mathrm{N}_2\mathrm{O}_4\) and brown \(\mathrm{NO}_2\) responds to restore balance by favoring the endothermic reaction that absorbs heat – thus, more \(\mathrm{NO}_2\) is formed, deepening the brown color observed.
Endothermic Reactions
Endothermic reactions are chemical processes that absorb energy from their surroundings, typically in the form of heat. This energy is necessary to break the bonds of the reactants, so the reaction will not proceed unless this energy is provided.

Visual signs, like the deepening of the brown color in the equilibrium of \(\mathrm{N}_2\mathrm{O}_4\) and \(\mathrm{NO}_2\), can indicate that a reaction is endothermic. In practice, endothermic reactions need a continuous input of energy to keep the reaction going, which could be in the form of thermal energy, as illustrated in the problem. As Le Chatelier's Principle suggests, increasing the temperature causes the endothermic reaction to be more favorable to absorb the excess heat, which is exactly why more \(\mathrm{NO}_2\) is produced when the flask is heated.
Change in Enthalpy
The change in enthalpy, designated by \(\Delta H\), represents the total heat change in a system during a chemical reaction at constant pressure. A positive \(\Delta H\) value signifies an endothermic reaction, indicating that the reaction absorbs heat from the surroundings. Conversely, an exothermic reaction has a negative \(\Delta H\), meaning it releases heat.

In the given exercise, the darkening of the brown color on heating implies that the system absorbs heat, pointing to a positive \(\Delta H\). This concept is crucial in the study of thermodynamics, as it helps chemists understand and predict how the equilibrium will shift in response to temperature changes or other external conditions. Knowing whether a reaction is endothermic or exothermic is essential for controlling industrial chemical processes and even for everyday cooking!

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Most popular questions from this chapter

Amounts of \(0.8 \mathrm{~mol}\) of \(\mathrm{PCl}_{5}\) and \(0.2\) mole of \(\mathrm{PCl}_{3}\) are mixed in a \(1 \mathrm{i}\) flask. At equilibrium, \(0.4\) mole of \(\mathrm{PCl}_{3}\) is present. The equilibrium constant for the reaction, \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) will be (a) \(0.05 \mathrm{~mol}^{-1}\) (b) \(0.13 \mathrm{~mol} 1^{-1}\) (c) \(0.013 \mathrm{~mol} 1^{-1}\) (d) \(0.60 \mathrm{~mol} 1^{-1}\)

For a gaseous equilibrium: \(2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}), K_{\mathrm{p}}\) has a value \(1.8\) at \(700 \mathrm{~K}\). The value of \(K_{c}\) for the equilibrium: \(2 \mathrm{~B}(\mathrm{~g})\) \(+\mathrm{C}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~A}(\mathrm{~g})\) at that temperature is about (a) \(0.031\) (b) 32 (c) \(57.4\) (d) \(103.3\)

In the equilibrium mixture: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\) \(\rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\), the mole ratio of gases are \(\sqrt{2}: \sqrt{2}: 10\), respectively. What would be the effect on the mole ratio on adding 5 mole of He gas at constant pressure? (a) No change (b) The new molar ratio becomes \(1: 1: 5\) (c) The new molar ratio becomes \(2: 2: 5\) (d) The new molar ratio becomes \(2: 2: 5 \sqrt{2}\)

\(\mathrm{XeF}_{6}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{XeOF}_{4}+2 \mathrm{HF}\) equilibrium constant \(=K_{1}\). \(\mathrm{XeO}_{4}+\mathrm{XeF}_{6} \rightleftharpoons \mathrm{XeOF}_{4}+\mathrm{XeO}_{3} \mathrm{~F}_{2}\) equilibrium constant \(=K_{2} .\) Then equilibrium constant for the following reaction will be: \(\mathrm{XeO}_{4}+2 \mathrm{HF} \rightleftharpoons \mathrm{XeO}_{3} \mathrm{~F}_{2}+\mathrm{H}_{2} \mathrm{O}\) (a) \(\frac{K_{1}}{K_{2}}\) (b) \(K_{1}+K_{2}\) (c) \(\frac{K_{2}}{K_{1}}\) (d) \(K_{2}-K_{1}\)

\(\mathrm{I}_{2}+\mathrm{I}^{-} \rightleftharpoons \mathrm{I}_{3}^{-} .\) This reaction is set up in aqueous medium. We start with \(1 \mathrm{~mol}\) of \(\mathrm{I}\), and \(0.5 \mathrm{~mol}\) of \(\mathrm{I}^{-}\) in \(1 \mathrm{~L}\) flask. After equilibrium is reached, excess of \(\mathrm{AgNO}_{3}\) gave \(0.25 \mathrm{~mol}\) of yellow precipitate. Equilibrium constant is (a) \(1.33\) (b) \(2.66\) (c) \(0.375\) (d) \(0.75\)

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