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Densities of diamond and graphite are \(3.5\) and \(2.4 \mathrm{~g} / \mathrm{ml}\), respectively. The increase in pressure (at the constant temperature) at the equilibrium in \(\mathrm{C}\) (diamond) \(\rightleftharpoons\) \(\mathrm{C}\) (graphite) will (a) favour the forward reaction (b) favour the backward reaction (c) have no effect (d) increases the equilibrium constant

Short Answer

Expert verified
Increasing pressure will favour the forward reaction because diamond, with a higher density, occupies less volume than graphite.

Step by step solution

01

Understanding Le Chatelier's Principle

Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This principle helps to predict the effects of changes in pressure, temperature, and concentration on a system at equilibrium.
02

Analyzing the Effect of Pressure on Equilibrium

For solids, the effect of pressure is related to their density. If the pressure is increased, the equilibrium will shift towards the phase with the higher density since this will result in a smaller volume. Between diamond and graphite, diamond has a higher density.
03

Determining the Direction of the Shift in Equilibrium

Since diamond has a higher density than graphite, increasing pressure at a constant temperature will favor the forward reaction, where Carbon as graphite transforms into Carbon as diamond. This is because the forward reaction results in a denser phase, which is associated with a smaller volume, and the system can reduce its volume to accommodate the increased pressure.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium refers to the state in a reversible chemical reaction where the rate of the forward reaction is equal to the rate of the backward reaction. In this balanced state, the amount of reactants and products remain constant over time, though they are not necessarily present in equal quantities.

For example, in the reversible reaction of carbon allotropes, diamond and graphite, the equilibrium represents the point at which diamond turns into graphite at the same rate graphite turns into diamond. This doesn't mean there are equal amounts of diamond and graphite but that their amounts are stable unless an external force is applied.

To understand equilibrium, it's essential to grasp that it's a dynamic process – molecules continuously convert from reactants to products and back, but their overall concentrations do not change unless the system is disturbed. This introduces the concept of Le Chatelier's Principle, a guideline for predicting how a change in conditions can shift the equilibrium point.
Pressure Effect on Equilibrium
Le Chatelier's Principle reveals critical insights into how equilibrium shifts in response to changes in pressure. This principle posits that if a system at equilibrium experiences an alteration in pressure, the equilibrium will move in the direction that negates the pressure change.

For reactions involving gases, an increase in pressure typically shifts the equilibrium towards the side with fewer moles of gas, as this reduces the overall volume and counteracts the increase in pressure. However, when dealing with solids, like diamond and graphite, the concept differs – the pressure effect is directly linked to the density of the solid phases.

With increased pressure, the system favors the formation of the denser phase, which occupies less volume. This is why in the example of carbon allotropes, a rise in pressure at constant temperature would encourage the process of graphite transforming into diamond – facilitating the phase with greater density and hence less volume. It's important to recognize that such pressure-induced equilibrium shifts do not alter the equilibrium constant, they simply reposition the equilibrium point.
Solid Phase Density
Density plays a pivotal role in predicting the outcome of pressure changes on a system at equilibrium involving solid phases. Density, defined as mass per unit volume, largely influences how the equilibrium shifts when external pressure is applied.

The density of a solid is a fixed property at a given temperature and pressure, determining how closely packed the molecules within the solid are. In context of the diamond and graphite equilibrium, diamond's density of 3.5 g/ml is higher than graphite's density of 2.4 g/ml, indicating that diamond's molecules are more compactly arranged than those in graphite.

Therefore, an increase in pressure favors the formation of the denser phase, namely diamond in this scenario, because it allows the system to occupy a smaller volume which is in line with Le Chatelier's Principle. This fundamental understanding of solid phase densities provides a clear explanation for why the process of diamond formation from graphite is preferred under high-pressure conditions.

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Most popular questions from this chapter

The concentration of a pure solid or liquid phase is not included in the expression of equilibrium constant because (a) solid and liquid concentrations are independent of their quantities. (b) solid and liquids react slowly. (c) solid and liquids at equilibrium do not interact with gaseous phase. (d) the molecules of solids and liquids cannot migrate to the gaseous phase.

Steam decomposes at high temperature according to the equation: \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) ; \quad \Delta H^{\circ}\) \(=240 \mathrm{~kJ} / \mathrm{mole}\) and \(\Delta S^{\circ}=50 \mathrm{JK}^{-1} / \mathrm{mole}\) The temperature at which the equilibrium constant \(\left(K_{\mathrm{p}}^{\circ}\right)\) becomes \(1.0\), is (a) \(4.8 \mathrm{~K}\) (b) \(4800 \mathrm{~K}\) (c) \(480 \mathrm{~K}\) (d) Impossible

For the reaction: \(2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) \(+\mathrm{Cl}_{2}(\mathrm{~g}), \Delta \mathrm{H}^{\circ}=18 \mathrm{kcal}\) and \(\Delta S^{\circ}=30 \mathrm{cal} / \mathrm{K}\) at \(300 \mathrm{~K}\). The equilibrium constant, \(K_{\mathrm{p}}^{\circ}\) of the reaction at \(300 \mathrm{~K}\) is (a) \(\mathrm{e}^{15}\) (b) \(\mathrm{e}^{-15}\) (c) \(\mathrm{e}^{-18}\) (d) \(\mathrm{e}^{-12}\)

When \(\alpha\) -D-glucose is dissolved in water, it undergoes a partial conversion to \beta-D-glucose. This conversion, called mutarotation, stops when \(64.0 \%\) of the glucose is in the \(\beta\) -form. Assuming that equilibrium has been attained, what is \(\Delta G^{\circ}\) for the reaction: \(\alpha\) -D-glucose \(\rightleftharpoons \beta-\mathrm{D}\) glucose, at this experimental temperature? (a) \(-R T \log _{10}(1.6)\) (b) \(-R T \log _{10}(1.78)\) (c) \(-R T \log _{e}(1.78)\) (d) \(-R T \log _{e}(1.6)\)

\(\mathrm{I}_{2}+\mathrm{I}^{-} \rightleftharpoons \mathrm{I}_{3}^{-} .\) 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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