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For a reaction, the value of \(K_{p}\) increases with increase in temperature. The \(\Delta H\) for the reaction would be : (a) positive (b) negative (c) zero (d) cannot be predicted

Short Answer

Expert verified
The \(\Delta H\) for the reaction would be positive.

Step by step solution

01

Understanding the Relationship between Kp and Temperature

For an equilibrium reaction, the relationship between the equilibrium constant (\(K_p\)) and temperature can be understood using Le Chatelier's principle and the van't Hoff equation. Le Chatelier's principle states that if a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the system will adjust itself to partially counteract the effect of the applied change. An increase in the value of the equilibrium constant with increasing temperature suggests that the reaction favors the production of products and thus is endothermic.
02

Applying the van't Hoff Equation

The van't Hoff isobaric equation relates the change in the equilibrium constant to the change in temperature: \(\frac{d\ln K}{dT} = \frac{\Delta H}{RT^2}\). An increase in \(K_p\) with temperature implies that \(\frac{d\ln K}{dT}\) is positive. Since \(R\) (the universal gas constant) and \(T^2\) (the square of the temperature) are always positive, the sign of \(\Delta H\) must be positive for the derivative to be positive.
03

Determining the Sign of ΔH

From the van't Hoff equation and the given information about the increase in \(K_p\) with temperature, it can be deduced that \(\Delta H\) must be positive. This is because an endothermic reaction (positive \(\Delta H\)) would absorb heat, causing the system to shift towards the products to counterbalance the increase in temperature, resulting in the observed increase in \(K_p\).

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Key Concepts

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

Le Chatelier's Principle
Imagine you're in a dance room balanced perfectly between dancers and sitters. Suddenly, the music tempo changes—what happens? Dancers either join in or sit down to restore the balance. This is the essence of Le Chatelier's principle, but in the world of chemistry.

When a chemical system at equilibrium experiences a change, such as a temperature increase, it adjusts to counter this. For example, if the temperature rises during a reaction, the system favors the reaction that absorbs heat to reduce the effect of this change—just like some dancers might sit down when the music gets too fast. This principle is vital in predicting how a system will respond to stress, ensuring students grasp the interconnectedness of chemical reactions and their reaction conditions.
Van't Hoff Equation
Chemistry often feels like a puzzle, where the van't Hoff equation is a critical piece that helps solve the mystery of how temperature and equilibrium are related. This powerful formula is like a secret code, \( \frac{d\ln K}{dT} = \frac{\Delta H}{RT^2} \), revealing how the equilibrium constant (K) of a reaction changes with temperature.

Think of it as a see-saw. If the temperature goes up, the equilibrium constant tilts in one direction. If the heat given off by the reaction is like the weight of children on the see-saw, the direction the equilibrium constant tilts (increases or decreases) tells you whether the reaction is absorbing heat (endothermic) or releasing it (exothermic). With this equation, students can predict the behavior of reactions under thermal stress.
Endothermic Reaction
An endotheric reaction is the chemical equivalent of a sponge soaking up water; it absorbs heat from its surroundings. Just like a sponge expands with water, the reaction's equilibrium positional shifts toward the products to accommodate the extra energy input due to an increase in temperature.

Imagine a cold pack, which gets cold when certain chemicals inside it absorb heat from the environment—that's an endothermic process in action! Understanding this concept is crucial for students, as it underpins many phenomena in both the lab and the real world, from the production of sports injury packs to photosynthesis in plants which fuels life on Earth.

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Most popular questions from this chapter

At \(273 \mathrm{~K}\) and 1 atm, 10 litre of \(\mathrm{N}_{2} \mathrm{O}_{4}\) decomposes to \(\mathrm{NO}_{2}\) according to equation $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ What is degree of dissociation \((\alpha)\) when the original volume is \(25 \%\) less than that of existing volume? (a) \(0.25\) (b) \(0.33\) (c) \(0.66\) (d) \(0.5\)

For which of the following reaction is product formation favoured by low pressure and low temperature? (a) \(\mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g) ; \quad \Delta H^{\circ}=172.5 \mathrm{~kJ}\) (b) \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}\) \(\Delta H^{\circ}=-21.7 \mathrm{~kJ}\) (c) \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\) \(\Delta H^{\circ}=-285 \mathrm{~kJ}\) (d) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) \(\Delta H^{\circ}=-541 \mathrm{~kJ}\)

At a certain temperature the equilibrium constant \(K_{c}\) is \(0.25\) for the reaction $$ A_{2}(g)+B_{2}(g) \rightleftharpoons C_{2}(g)+D_{2}(g) $$ If we take 1 mole of each of the four gases in a 10 litre container, what would be equilibrium concentration of \(A_{2}(g) ?\) (a) \(0.331 \mathrm{M}\) (b) \(0.033 M\) (c) \(0.133 \mathrm{M}\) (d) \(1.33 \mathrm{M}\)

The equilibrium constant \(K_{p}\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ \(4.0\) at \(1660^{\circ} \mathrm{C}\). Initially \(0.80\) mole \(\mathrm{H}_{2}\) and \(0.80\) mole \(\mathrm{CO}_{2}\) are injected into a \(5.0\) litre flask. What is the equilibrium concentration of \(\mathrm{CO}_{2}(g) ?\) (a) \(0.533 \mathrm{M}\) (b) \(0.0534 \mathrm{M}\) (c) \(0.535 \mathrm{M}\) (d) None of these

A catalyst increases the rate of a reaction by: (a) increasing the activation energy of the reaction (b) increasing the value of rate constants \(\left(k_{f}\right.\) and \(k_{b}\) ) (c) increasing the enthalpy change of the reaction (d) decreasing the enthalpy change of the reaction

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