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The equilibrium constant of a reaction is \(300 .\) If the volume of reaction flask is tripled, the equilibrium constant is: (a) 300 (b) 600 (c) 900 (d) 100

Short Answer

Expert verified
The equilibrium constant remains 300, so the answer is (a) 300.

Step by step solution

01

Understanding Equilibrium Constant

The equilibrium constant, denoted as \(K\), is a measure of the ratio of concentrations of products to reactants at equilibrium. It is important to understand that the value of \(K\) is dependent on temperature for a given reaction.
02

Effect of Volume Change on Equilibrium Constant

The equilibrium constant \(K\) is primarily dependent on temperature and not on the volume or pressure unless the reaction involves gases. Volume change doesn't alter \(K\) itself. It affects the position of equilibrium only if there is a difference in the moles of gaseous products and reactants.
03

Analyzing the Problem

Since the problem states that the equilibrium constant of the reaction is \(300\), and it asks for the equilibrium constant if the volume is tripled, we need to recognize that since \(K\) is not affected by volume change, \(K\) remains the same.
04

Conclusion

Based on the understanding that equilibrium constant \(K\) does not change with volume alteration, the equilibrium constant of the reaction will still be \(300\) regardless of the tripling of the volume.

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Key Concepts

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

Effect of volume change on equilibrium
When a chemical reaction is at equilibrium, the amounts of reactants and products reach a balance. However, if the volume of the reaction container changes, this can affect the equilibrium position, especially in reactions involving gases. For such reactions, a change in volume changes the pressure according to the ideal gas law. This change can cause the system to shift its equilibrium to counteract the pressure change if there's a disparity in the number of moles of gaseous reactants and products.
  • If the volume increases, pressure decreases, prompting the system to shift toward the side with more moles of gas, thus partially increasing pressure.
  • Conversely, if the volume decreases, the pressure increases, and the equilibrium shifts toward the side with fewer moles of gas, which decreases pressure.
Despite influencing the equilibrium position, altering volume doesn't affect the equilibrium constant, since it's solely a function of temperature.
Equilibrium in chemical reactions
Chemical equilibrium refers to the state where the forward and reverse reaction rates are equal, meaning the concentrations of reactants and products remain constant over time. This doesn't mean that the amounts are equal, just that their rates of formation and consumption are balanced. At this point, the system is dynamically stable.
Equilibrium is described by the equilibrium constant, denoted as K, which quantifies the relative concentrations of reactants and products at equilibrium. It's calculated using the formula:\[ K = \frac{{[ ext{Products}]^m}}{{[ ext{Reactants}]^n}} \]where m and n are the stoichiometric coefficients.
The value of K provides insight into the reaction's favorability:
  • If K is much larger than 1, the equilibrium largely favors products.
  • If K is much less than 1, the reactants are favored.
  • An intermediate K value suggests significant concentrations of both reactants and products.
Understanding equilibrium is key to manipulating chemical reactions in practical applications, such as in industrial synthesis processes.
Temperature dependence of equilibrium constant
The equilibrium constant is sensitive to temperature changes, influencing the direction either the forward or reverse reaction. This is described by the Van't Hoff equation, which shows that K changes as temperature changes. \[ \ln(\frac{K_2}{K_1}) = -\frac{\Delta H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]where:
  • \( \Delta H^\circ \) is the standard enthalpy change.
  • \( R \) is the gas constant.
  • \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.
The value of \( \Delta H^\circ \) determines how the equilibrium position alters with temperature:
  • If \( \Delta H^\circ > 0 \), the reaction is endothermic, and K increases with temperature.
  • If \( \Delta H^\circ < 0 \), the reaction is exothermic, and K decreases with temperature.
This temperature dependence is crucial for designing and controlling chemical reactions in conditions where temperature cannot remain constant.

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Most popular questions from this chapter

For a gaseous equilibrium: \(2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}), \mathrm{K}_{\mathrm{p}}\) has a value of \(1.8\) at \(700 \mathrm{~K} .\) What is the value of \(\mathrm{K}_{\mathrm{c}}\) for the equilibrium \(2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~A}\) at the same pressure: (a) \(0.031\) (b) \(1.3 \times 10^{-3}\) (c) \(44.4\) (d) 38

Which of these reactions are having homogeneous equilibrium: (a) \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}\) \(' \mathrm{~g}^{\prime} \quad{ }^{6} \mathrm{~g}^{\prime}\) (b) \(\mathrm{H}_{2}+\mathrm{Cl}_{2} \rightleftharpoons 2 \mathrm{HCl}\) \(' \mathrm{~g}^{\prime} \quad{ }^{2} \mathrm{~g}^{\prime} \quad{ }^{\mathrm{s}} \mathrm{g}^{\prime}\) (c) \(\mathrm{CaCO}_{3} \rightleftharpoons \mathrm{CaO}+\mathrm{CO}_{2}\) 's' \(\quad{ }^{\prime} \mathrm{s}^{\prime} \quad{ }^{\prime} \mathrm{g}\) ' (d) Both (a) and (b)

In a reaction \(\mathrm{A}_{2}(\mathrm{~g})+4 \mathrm{~B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}_{4}(\mathrm{~g}) ; \Delta \mathrm{H}<0\) The formation of \(\mathrm{AB}_{4}(\mathrm{~g})\) will be favoured by: (a) Low temperature and high pressure (b) High temperature and high pressure (c) Low temperature and low pressure (d) High temperature and low pressure

The equilibrium constant for the reaction: \(\mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) is \(K_{c}=4.9 \times 10^{-2}\). the value of \(K\) for the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) will be (a) 416 (b) \(2.40 \times 10^{-3}\) (c) \(9.8 \times 10^{-2}\) (d) \(4.9 \times 10^{-2}\)

The vapour density of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at a certain temperature is 30\. What is the percentage dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at this temperature: (a) \(53.3\) (b) \(106.6\) (c) \(26.7\) (d) None of these

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